if y is given and you need to find third derivative of y (given as y'''), what are the steps: explain what one needs to do and say it in your words.

Answer:

Step-by-step explanation:

D

Answer:

Step-by-step explanation:

We'Re looking at a normal distribution here- let's start by drawing it out to the mean me- is 315 grams standard. Deviation is 16 point. We want to know the weight corresponding to each of these events and we can use either the appendix which i assume is a z, school table or excel so the first 1. We want the highest 20 percent up here somewhere. This is what we would call the 80 percent. It separates the bottom 80 percent from the highest 20 percent. So how do we work the well? We need to start by getting the z score for it. So how do we get that, while in excel you're going to use the norm inverse function which looks like this? So it's calls norm and then in here you just put in x, where x is your percentage and that will spit out the z score and i'm using this rather than a table, because it will give me a more exact value. So we're going to do that, and so here the percent is the 80, so it's not .8 and that spits out the z score of nort .8416 to 4 decimal places. But i'm always going to this exact score, because we now have to turn it from a z score to a piece of real data. Z score is a measure of how many standard deviations away from the mean a value is so we're. Looking for the value not .84 standard deviations, above the mean or if we write it like this x, is equal to z, sigma plus mu. So here we take our exact zedscore, because we can still just use excels, multiply it by 16 and add it on to the mean and we'll get our value of 328.447328 .47 grams to 2 decimal places for part b. We want to be middle 60 percent. Now we need to cut off points and within here where, in this interval we have 60 percent of values, which means we have 40 percent of values, not in here. So we can label our 3 sections. These 2 add up to 40 percent, so they have to be 20 percent. Each are called nor .2 and not .2, and then this middle bit here is nor .6 for the total of 100 percent point. So we need these cut off points when you, the z, scores first, and because these are equal distances away from a mine they're going to have the same z score. Just 1 is going to be positive or negative, so z is going to be equal to plus and minus. Let'S look at the lower 1. This is the 20 percent here so into this excel command. We put 20 percent nor .2 and out of it we get the z score of minus, not .8416. So it's actually very similar to the top question, because the top question asked you for the 80 percent be 80 percent. Is the upper cut off point here? 20? Is below cutoff point, so we already have the up 1, let's just calculate below 1, so we've got to be minus, nor .8416 multiplied by 16, but the standard deviation and on to the mean- and we get 301.53 and the upper cut off point is from part A so that's the middle 60 percent makes more space a part c. We want to be highest 80 percent. So now what we want is the cut off between the lowest 20 and the highest 80, which we've just got from part c part b. It'S this lower 1, here, 301.53 grams. That'S an easy! 1! Now we want the lowest 15 percent, so the lowest 15 percent is the 15 percent. So we go to our exylgamant put in the 15 for percent, so that would be no .15 and it's a z score of minus 1.036 keeping the exact value put into this formula. We multiply our z by 16, as on 315, to get 298.42 grams.

The weight that corresponds to this event are approximately 344.03 grams and 375.97 grams.

To find the weight that corresponds to each event, we need to use the standard normal distribution, which has a mean of 0 and a standard deviation of 1. We can convert the given mean and standard deviation to z-scores using the formula:

z = (x - μ) / σ

where x is the weight we want to find, μ is the mean (360 grams), and σ is the standard deviation (9 grams).

Then, we can use a standard normal distribution table or calculator to find the probability of each event, and convert it back to a weight using the inverse of the z-score formula:

x = μ + z * σ

where z is the z-score that corresponds to the desired probability.

Event 1: The weight is less than 345 grams.

z = (345 - 360) / 9 = -1.67

Using a standard normal distribution table or calculator, we find that the probability of a z-score less than -1.67 is approximately 0.0475.

x = 360 + (-1.67) * 9 = 344.03 grams

Therefore, the weight that corresponds to this event is approximately 344.03 grams.

Event 2: The weight is between 355 and 365 grams.

First, we need to find the z-scores that correspond to the two boundaries:

z1 = (355 - 360) / 9 = -0.56

z2 = (365 - 360) / 9 = 0.56

Using a standard normal distribution table or calculator, we find that the probability of a z-score less than -0.56 is approximately 0.2123, and the probability of a z-score less than 0.56 is approximately 0.7123. Therefore, the probability of a z-score between -0.56 and 0.56 is:

0.7123 - 0.2123 = 0.5

x1 = 360 + (-0.56) * 9 = 355.16 grams

x2 = 360 + (0.56) * 9 = 364.84 grams

Therefore, the weight that corresponds to this event is any weight between 355.16 and 364.84 grams.

Event 3: The weight is greater than 375 grams.

z = (375 - 360) / 9 = 1.67

Using a standard normal distribution table or calculator, we find that the probability of a z-score greater than 1.67 is approximately 0.0475.

x = 360 + (1.67) * 9 = 375.97 grams

Therefore, the weight that corresponds to this event is approximately 375.97 grams.

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(a) The interval (-∞, ∞).

(b) The interval (-∞, ∞).

(c) The interval (-∞, ∞).

(d) The interval (-π/2, π/2) \ {0}.

(a) The longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution is the interval where the coefficient function, 3t, is continuous and bounded. Since 3t is a continuous and bounded function for all t in the interval (-∞, ∞), the given initial value problem is certain to have a unique twice-differentiable solution for all t in (-∞, ∞).

(b) The longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution is the interval where the coefficient functions, t(t - 4), 3t, and 4, are continuous and bounded. Since t(t - 4), 3t, and 4 are continuous and bounded functions for all t in the interval (-∞, ∞), the given initial value problem is certain to have a unique twice-differentiable solution for all t in (-∞, ∞).

(c) The longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution is the interval where the coefficient functions, cost and In|t|, are continuous and bounded. Since cost and In|t| are continuous and bounded functions for all t in the interval (-∞, ∞), the given initial value problem is certain to have a unique twice-differentiable solution for all t in (-∞, ∞).

(d) The longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution is the interval where the coefficient functions, x - 2, 1, and (x - 2)tanx, are continuous and bounded. Since x - 2, 1, and (x - 2)tanx are continuous and bounded functions for all x in the interval (-π/2, π/2) \ {0} , the given initial value problem is certain to have a unique twice-differentiable solution for all x in (-π/2, π/2) \ {0}.

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The given question is incomplete, the complete question is:

determine the longest interval in which the given initial value problem is certain to have a unique twice- differentiable solution. Do not attempt to find the solution. (a) ty" + 3y = 1, y(1) = 1, y'(1) = 2   (b)   t(t – 4)y" + 3ty' + 4y = 2, y(3) = 0, y'(3) = -1 (c)   y" + (cost)y' + 3( In |t|) y = 0, y(2) = 3, y'(2) = 1 (d) (x - 2)y"+y' +(x - 2)(tan x) y = 0, y(3) = 1, y'(3) = 2

To find the area of the triangle with vertices (9,-1), (6,1), and (6,3), we can use the formula:

[tex]$A = \frac{1}{2} \left| x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2) \right|$[/tex]

where (x1, y1), (x2, y2), and (x3, y3) are the coordinates of the three vertices.

Plugging in the coordinates, we get:

[tex]$A = \frac{1}{2} \left| 9(1-3) + 6(3-(-1)) + 6((-1)-1) \right|$[/tex]

[tex]$A = \frac{1}{2} \left| -6 + 24 - 12 \right| = \frac{1}{2} \cdot 6 = 3$[/tex]

Therefore, the area of the triangle is 3 square units.

To find the area of the triangle with vertices (0,-8), (0,-10), and (7,-10), we can again use the formula:

[tex]$A = \frac{1}{2} \left| x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2) \right|$[/tex]

Plugging in the coordinates, we get:

[tex]$A = \frac{1}{2} \left| 0((-10)-(-10)) + 0((7)-0) + 7((-8)-(-10)) \right|$[/tex]

$A = \frac{1}{2} \cdot 14 = 7$

Therefore, the area of the triangle is 7 square units.

To find the area of the triangle with vertices (6,-7), (3,-1), and (-1,4), we can again use the formula:

[tex]$A = \frac{1}{2} \left| x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2) \right|$[/tex]

Plugging in the coordinates, we get:

[tex]$A = \frac{1}{2} \left| 6((-1)-4) + 3(4-(-7)) + (-1)((-7)-(-1)) \right|$[/tex][tex]$A = \frac{1}{2} \cdot 55 = \frac{55}{2}$[/tex]

Therefore, the area of the triangle is $\frac{55}{2}$ square units.

To find the area of the quadrilateral with vertices (-6,1), (-9,1), (-6,-4), and (-9,-4), we can divide it into two triangles and find the area of each triangle using the determinant method. The area of the quadrilateral is the sum of the areas of the two triangles.

First, we find the coordinates of the diagonals:

$D_1=(-6,1)$ and $D_2=(-9,-4)$

The area of the quadrilateral can be calculated as:

\begin{align*}

\text{Area}&=\frac{1}{2}\left|\begin{array}{cc} x_1 & y_1 \ x_2 & y_2 \end{array}\right| + \frac{1}{2}\left|\begin{array}{cc} x_2 & y_2 \ x_3 & y_3 \end{array}\right|\

&=\frac{1}{2}\left|\begin{array}{cc} -6 & 1 \ -9 & -4 \end{array}\right| + \frac{1}{2}\left|\begin{array}{cc} -9 & -4 \ -6 & -4 \end{array}\right|\

&=\frac{1}{2}\cdot 21 + \frac{1}{2}\cdot 9\

&=\frac{15}{2}\

\end{align*}

Therefore, the area of the quadrilateral is $\frac{15}{2}$ square units.

To find the area of the pentagon with vertices (0,3), (-3,3), (-5,1), (-3,-3), and (-1,-2), we can divide it into three triangles and find the area of each triangle using the determinant method. The area of the pentagon is the sum of the areas of the three triangles.

First, we find the coordinates of the diagonals:

$D_1=(0,3)$ and $D_2=(-1,-2)$

The area of the pentagon can be calculated as:

\begin{align*}

\text{Area}&=\frac{1}{2}\left|\begin{array}{cc} x_1 & y_1 \ x_2 & y_2 \end{array}\right| + \frac{1}{2}\left|\begin{array}{cc} x_2 & y_2 \ x_3 & y_3 \end{array}\right| + \frac{1}{2}\left|\begin{array}{cc} x_3 & y_3 \ x_4 & y_4 \end{array}\right|\

&=\frac{1}{2}\left|\begin{array}{cc} 0 & 3 \ -3 & 3 \end{array}\right| + \frac{1}{2}\left|\begin{array}{cc} -3 & 3 \ -5 & 1 \end{array}\right| + \frac{1}{2}\left|\begin{array}{cc} -5 & 1 \ -3 & -3 \end{array}\right|\

&=\frac{1}{2}\cdot 9 + \frac{1}{2}\cdot (-6) + \frac{1}{2}\cdot (-8)\

&=\frac{5}{2}\

\end{align*}

Therefore, the area of the pentagon is $\frac{5}{2}$ square units.

Area of triangle whose vertices are (6,1), (9,-1) and (6,-3) is 6 square units and the area of triangle whose vertices are (0,-8), (7,-10) and (0,-10) is 7 square units.

A polygon having 3 edges and 3 vertices is called a triangle. It is one of the fundamental geometric forms.

Lets find the area of triangle ( Pink Colour) whose vertices are (6,1), (9,-1) and (6,-3), [tex]Area = \frac{1}{2} [x_{1}(y_{2} -y_{3}) + x_{2}(y_{3}-y_{1}) + x_{3}(y_{1}-y_{2} ) ][/tex]

Area = 1/2 [ 6 ( -1 - (-3) ) + 9( -3 -1 ) + 6( 1 - ( -1 ) ) ]

Area = 1/2 [6 * 2 + 9 * (-4) + 6 * 2]

Area = 1/2 [12-36+12] = 1/2 (-12) = -6

Therefore , Area of Triangle is 6 square units.

Now, Lets find the area of triangle ( Brown Colour ) whose vertices are (0,-8), (7,-10) and (0,-10),

[tex]Area = \frac{1}{2} [x_{1}(y_{2} -y_{3}) + x_{2}(y_{3}-y_{1}) + x_{3}(y_{1}-y_{2} ) ][/tex]

Area = 1/2 [  0( -10 - ( -10 )) + 7 ( -10 - ( -8 ) ) + 0 ( -8 - ( -1- ) ) ]

Area = 1/2 [ 0 + 7 * (-2) + 0]

Area = 1/2 ( -14 ) = -7

Therefore, Area of Triangle is 7 square units.

Now. Lets find the area of Rectangle( Blue Colour ) whose length is 5 unit and Breadth is 3 unit.

So, Area of Rectangle = Length * Breadth

= 5 * 3 square units

= 15 square units.

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Answer:

We can use the formula |a × b| = |a| |b| sin θ to solve for the magnitude of the cross product |a × b|, where θ is the angle between vectors a and b. In this case, we have |a × b| = 1 and θ = π/4, so we can write:

1 = |a| |b| sin(π/4)

Simplifying, we have:

|a| |b| = √2

Now, we need to find the dot product a · b. We know that:

a · b = |a| |b| cos θ

where θ is the angle between vectors a and b. Since we're given the angle between a and b, we can substitute θ = π/4 and use the value we found for |a| |b|:

a · b = (√2) cos(π/4) = (√2)/2

Therefore, a · b is equal to (√2)/2.

Step-by-step explanation:

Answer: The poster should have width 17.50 CM

Step-by-step explanation:

Given that boasted I have a total area 245 cm square area of a poster is 200 and 45 20 m square. And it is in the rectangle format. So we know that the poster is always in the rectangle format and the area of rectangular area is equal to L N T W. So 245 will be equal to Ln tW. From this. We need to find L. So L is equal to 245 divided by W. Now consider the diplomatic representation of the poster. So it has mentioned that there is a margin around the edges of six cm at the top. This is 6cm and four cm at the sites. All the four sides 3 sides are four cm. So from this we need to find land and the doctor posted area that is printed area. The first wine printed with www. Z. Quilter. Now let this be total birth will be W. And this will be four and this will be four. Therefore posted with will be we need to find this part alone. So W -4 -4 will give the this part with. So W -4 -4. So post printed with PW will be equal to W -8. Similarly printed lunch will be equal to The total length is already we have found 245 by W. And we need to find this part length. So we have to subtract six and four from the total length so that that will give them this part length, So -6 -4. So printed length will be equal there 245 Divided by W -10. And we know that formula for area of a rectangle. Urz D is equal to L W. Now substitute the printed with and printed length in the area formula. We have to find the printed area. So Printed area Zeke Walter W -8 into 245 Divided by W -10. To simplify this, we get 325 minus 10. W -100 and 30,960 W. to the power of -1. Now fine D A by D. T. Not D D. This is D. W. So this is equal to differentiation of constant alma zero, this is minus 10 plus 900 and 60 W. to the power of -2 and equate this to be equal to zero. We out to find the maximum width so D A by D W is equal to zero, therefore minus 10 1960 W. To the power of -2 will be equal to zero. To simplify this, we get them maximum with value the W is equal term I wrote off 196. Therefore the value of W. Z quilter plus or minus 14. We will neglect the negative values since we cannot be negative. So one we assume the positive values. So what will be equal to 14 and length will be equal to 245 divided by 14, So which will be equal to 17.50 centimeters. And they have concluded that At W is equal to 14 cm and lunch will be equal to 17.50 cm needed to print the largest area. I hope you found the answer to school. Thank you.

The cardinality of set from the given vein diagram is found as 2.

Think about set A. The set A is said to be finite and so its cardinality is same to the amount of elements n if it includes precisely n items, where n  ≥  0. |A| stands for the cardinality of such a set A.

It turns out that there are two kinds of infinite sets that we need to determine between since one form is much "bigger" than the other. Particularly, one type is referred to as countable and the other as uncountable.

From the given figure

Set A = {8 , 8, 3, 6}

Compliment of Set B (elements not present in set B):

Set [tex]B^{c}[/tex] = {8, 8, 6(pink), 3(white)}

Thus,

(A∩ [tex]B^{c}[/tex] ) = {8, 8} (present in both set)

n (A∩ [tex]B^{c}[/tex] ) = 2 (cardinal number)

Thus, the cardinality of the set from the given vein diagram is found as 2.

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Answer:

23.8 cm

Step-by-step explanation:

17 * 140% = 17 * 1.4 = 23.8 cm

The image of the house would be 23.8 cm tall on the screen.

To calculate the height of the image of the house on the screen, we can use the given zoom factor of 140%.

The zoom factor of 140% means that the images on the screen are 140% as long and 140% as wide compared to when they are printed on a sheet of paper.

To calculate the height of the image on the screen, we need to multiply the printed height by the zoom factor (140% or 1.4).

Height on the screen = Printed height * Zoom factor

Height on the screen = 17 cm * 1.4

Height on the screen = 23.8 cm

Therefore, the image of the house would be 23.8 cm tall on the screen.

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Answer:

The unit circle has many different angles that each have a corresponding point on the circle. The coordinates of each point give us a way to find the tangent of each angle. The tangent of an angle is equal to the y-coordinate divided by the x-coordinate.

For the first question, the equation for the line is y = -x + 11. This comes from the fact that the slope for a line perpendicular to the line x = 5 is -1. From there, we can use the point (5,6) to calculate the y-intercept, which is 11.

For the second question, the equation for the line is y = 7x. This comes from the fact that the slope for a line parallel to the line 7x - y = -7 is 7. Since the point (0,0) is already on the line, the equation is already solved.

For the third question, the equation for the line is x = 8. This comes from the fact that the slope for a line with an undefined slope is 0. Since the point (8,2) is already on the line, the equation is already solved.

Answer:

B) Group 2

Step-by-step explanation:

Group 1 wrote it correctly because there are 3 groups of 4x and 3 groups of 3

Group 2 wrote it incorrectly because that would mean 9 groups of 4x, which is not the case here

Group 3 wrote it correctly because 3(4x) + 3(3) = 3(4x+3) due to the Distributive Property

Group 4 wrote it correctly because it is an expansion of Group 3's expression

Answer: $27.98

Step-by-step explanation:

22.00 × .2= 4.40

22 + 4.40 = 26.40

26.40 × .06 = 1.584

26.40 + 1.584 = 27.984

Round to the nearest hundred so the total paid by the customer would be 27.98

The rental cost for each movie and each video game is $3.5 and $5.5 respectively.

Let

cost of each movie = x

Cost of each video game = y

5x + 3y = 34

2x + 12y = 73

Multiply (1) by 4

20x + 12y = 136

2x + 12y = 73

subtract the equations to eliminate y

18x = 63

divide both sides by 18

x = 63/18

x = 3.5

Substitute x = 3.5 into (1)

5x + 3y = 34

5(3.5) + 3y = 34

17.5 + 3y = 34

3y = 34 - 17.5

3y = 16.5

y = 16.5/3

y = 5.5

Therefore, $3.5 and $5.5 is the rental cost of each movie and video game respectively.

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We can see here the values of the interior angles will be: A) m∠X = 131º, m∠Y = 16º, m∠Z = 33º.

An interior angle is an angle created between two adjacent sides of a polygon. To put it another way, it is the angle created by two polygonal sides that have a shared vertex.

Sum of interior angles of a triangle = 180°

[tex]2p + \frac{1}{4} p + \frac{1}{2} p = 180[/tex]

11p/4 = 180°

p = 720°/11

m∠X = 2p = 2 ×  720°/11 = 130.9 ≈ 131°

m∠Y = [tex]\frac{1}{4} p[/tex] = 1/4 × 720°/11 = 16.3 ≈ 16°

m∠Z = [tex]\frac{1}{2} p[/tex] = 1/2 × 720°/11 = 32.7 ≈ 33°

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The inequality is X > 0 and a word sentence represent the graph is  X  the graph of a number line with an open circle on zero and an arrow pointing to the right.

The inequality X > 0 represents the graph of a number line with an open circle on zero to left and an arrow pointing to the right. This means that any value of X that is greater than zero is a valid solution for the inequality.

In other words, X can be any positive number, such as 1, 2, 3, and so on. However, X cannot be zero or any negative number, as those values do not satisfy the inequality. Therefore, the word sentence that represents this inequality is "X is greater than zero."

This means that X must be a positive number, and it can be any value that is greater than zero.

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Answer:

1. x = 11

Step-by-step explanation:

1. x + 7 = 18

move 7 to right then change the sign
        x = 18 - 7
        x = 11
2. x - 13 = 15

move -13 to right then change the sign
        x = 15 + 13
        x = 28
3. 8x = 64
   8      8
divided by 8 both side
        x = 8
4. 5x-13-12=0 it this the correct given?
add same variable
   5x = 13 + 12
   5x = 25
   5      5

divided by 5 both side
         x = 5

Answer:

E.  x² + (y - 3)² = 16

Step-by-step explanation:

The equation of a circle in the standard (x, y) coordinate plane with center (h, k) and radius r is given by:

[tex]\boxed{(x - h)^2 + (y - k)^2 = r^2}[/tex]

To find the equation of the circle with a radius of 4 units and the same center as the circle determined by x² + y² - 6y + 4 = 0, we need to first write the equation of the second circle in the standard form.

We can complete the square for y to rewrite this equation in standard form. To do this move the constant to the right side of the equation:

[tex]\implies x^2 + y^2 - 6y + 4 = 0[/tex]

[tex]\implies x^2 + y^2 - 6y = -4[/tex]

Add the square of half the coefficient of the term in y to both sides of the equation:

[tex]\implies x^2 + y^2 - 6y +\left(\dfrac{-6}{2}\right)^2= -4+\left(\dfrac{-6}{2}\right)^2[/tex]

[tex]\implies x^2 + y^2 - 6y +9= -4+9[/tex]

[tex]\implies x^2 + y^2 - 6y +9=5[/tex]

Factor the perfect square trinomial in y:

[tex]\implies x^2+(y-3)^2=5[/tex]

[tex]\implies (x-0)^2 + (y-3)^2=5[/tex]

So the center of this circle is (0, 3) and its radius is √5 units.

Since the new circle has the same center, its center is also (0, 3).

We know its radius is 4 units, so we can write the equation of the new circle as:

[tex]\implies (x - 0)^2 + (y - 3)^2 = 4^2[/tex]

[tex]\implies x^2 + (y - 3)^2 = 16[/tex]

Therefore, the equation of the circle in the standard (x, y) coordinate plane with a radius of 4 units and the same center as the circle determined by x² + y² - 6y + 4 = 0 is x² + (y - 3)² = 16.

To find:-

Answer:-

The given equation of the circle is ,

[tex]\implies x^2+y^2-6y + 4 = 0 \\[/tex]

Firstly complete the square for y in LHS of the equation as ,

[tex]\implies x^2 + y^2 -2(3)y + 4 = 0 \\[/tex]

Add and subtract 3² ,

[tex]\implies x^2 +\{ y^2 - 2(3)(y) + 3^2 \} -3^2 + 4 = 0 \\[/tex]

The term inside the curly brackets is in the form of a²-2ab+b² , which is the whole square of "a-b" . So we may rewrite it as ,

[tex]\implies x^2 + (y-3)^2 -9 + 4 = 0 \\[/tex]

[tex]\implies x^2 + (y-3)^2 - 5 = 0 \\[/tex]

[tex]\implies x^2 + (y-3)^2 = 5\\[/tex]

can be further rewritten as,

[tex]\implies (x-0)^2 + (y-3)^2 = \sqrt5^2\\[/tex]

now recall the standard equation of circle which is ,

[tex]\implies (x-h)^2 + (y-k)^2 = r^2 \\[/tex]

where,

So on comparing to the standard form, we have;

[tex]\implies \rm{Centre} = (0,3)\\[/tex]

Now we are given that the radius of second circle is 4units . On substituting the respective values, again in the standard equation of circle, we get;

[tex]\implies (x-h)^2 + (y-k)^2 = r^2 \\[/tex]

[tex]\implies (x-0)^2 + (y-3)^2 = 4^2 \\[/tex]

[tex]\implies \underline{\underline{\red{ x^2 + (y-3)^2 = 16}}}\\[/tex]

and we are done!

Therefore, the scale factor of the dilation of the shape is 2.

A scale factor is a ratio that describes the proportional relationship between two similar figures. It represents how much larger or smaller one figure is compared to the other, and it is calculated by dividing a corresponding measurement (such as side length, perimeter, or area) of the larger figure by the corresponding measurement of the smaller figure. Scale factor is used in mathematics, particularly in geometry and measurement, to describe the transformation of one figure into another through dilation or resizing. It is represented by a number or a ratio, such as 2:1 or 1/2, which indicates how many times larger or smaller the new figure is compared to the original.

Here,

In the given picture, it appears that the distance from the center of dilation (the origin) to the pre-image (the original figure) is 4 units, and the distance from the center of dilation to the image (the transformed figure) is 8 units. The scale factor of the dilation is equal to the ratio of the distance from the center of dilation to the image and the distance from the center of dilation to the pre-image.

So, the scale factor of the dilation is:

8 units ÷ 4 units = 2

Therefore, the scale factor of the dilation is 2.

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Answer:

x = 2

y = -3/4

Step-by-step explanation:

1. Substitute y=1/8x -1 in −5x+4y=−13

-5x+4(1/8x -1) = -13

2. Solve for x

-5x + 4/8x - 4 = -13

-9/2x - 4 = -13

-9/2x = -9

x = 2

3. Now that you know x = 2, plug it into y=1/8x - 1 to find what y is.

y= 1/8(2) - 1

y= 2/8 - 1

y= -3/4

Answer:

Step-by-step explanation:

t+t+t= 3t

3t = 12

12/3=t

4=t

Answer:

Step-by-step explanation:

5m(4m) = 20m^2

The area of the enlarged logo will be 9 times larger than the original logo.

When an object is enlarged or scaled up how does it area change ?

When an object is enlarged or scaled up by a factor of [tex]k[/tex], both its length and width are multiplied by [tex]k[/tex]. Therefore, the new length is [tex]k[/tex] times the original length, and the new width is [tex]k[/tex] times the original width.

The area of the new object is the product of the new length and width, which is ([tex]k[/tex] times the original length) multiplied by ([tex]k[/tex] times the original width), or [tex]k^2[/tex] times the original area.

Therefore, the area of an object increases by a factor of [tex]k^2[/tex] when the object is enlarged or scaled up by a factor of [tex]k[/tex].

Calculating how many times larger the area of the enlarged logo will be :

The original logo has dimensions of 3 inches by 5 inches, so its area is 3 x 5 = 15 square inches.

Mr. James wants to enlarge the logo so that the dimensions are 3 times larger than the original. This means the new dimensions will be 9 inches by 15 inches.

To determine how many times larger the area of the enlarged logo will be, we need to compare the areas of the original logo and the enlarged logo. The area of the enlarged logo is 9 x 15 = 135 square inches.

To find out how many times larger the area of the enlarged logo is compared to the original logo, we divide the area of the enlarged logo by the area of the original logo:

135 square inches ÷ 15 square inches = 9

Therefore, the area of the enlarged logo will be 9 times larger than the area of the original logo.

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The appropriate inference from the data is (B) Since [tex]20%[/tex] of this group read below grade level, we cannot draw any generalizations about the population. Thus, option B is correct.

A representative sample is a subset of data, often drawn from a wider population, that can show qualities that are similar.

Because the data produced contains more manageable, smaller representations of the larger group, representative sampling aids in the analysis of bigger groups.

Although the sample may be representative of seventh-grade American students, it is not necessarily representative of all seventh-graders or all American children. Hence, without additional data or research, we cannot extrapolate the sample's results to the overall population.

Therefore, 20%20, percent of all American students in seventh grade were reading below grade level.

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The equation of the line that is parallel to the given line and passes through the point (-3, 2) is 3x - 4y = -17.

If the line's equation is axe + by + c = 0 and the coordinates are (x1, y1).

To find the equation of a line parallel to a given line, we must first understand that parallel lines have the same slope. As a result, we must first determine the slope of the given line.

3x - 4y = -17 is the given line. To determine its slope, solve for y and write the equation in slope-intercept form:

-3x - 4y = -3x - 17 y = (3/4)x + (17/4)

This line has a 3/4 slope.

Now we want to find the equation of a parallel line that passes through the point (-3, 2). Because the new line is parallel to the given line, it has a slope of 3/4.

We can write the equation of the new line using the point-slope form of the equation of a line as:

y - y1 = m(x - x1)

where m represents the slope and (x1, y1) represents the given point (-3, 2).

When m = 3/4, x1 = -3, and y1 = 2, we get:

y - 2 = (3/4)(x - (-3))

y - 2 = (3/4)(x + 3)

Divide both sides by 4 to get rid of the fraction.

4y - 8 = 3x + 9

3x - 4y = -17

As a result, the equation of the parallel line that passes through the point (-3, 2) is 3x - 4y = -17.

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The function rule for this graph is  y = -1/2(x) + 2.

In Mathematics, the point-slope form of a straight line can be calculated by using the following mathematical expression:

y - y₁ = m(x - x₁) or [tex]y - y_1 = \frac{(y_2- y_1)}{(x_2 - x_1)}(x - x_1)[/tex]

Where:

At data point (0, 2), a linear equation in slope-intercept form for this line can be calculated by using the point-slope form as follows:

[tex]y - y_1 = \frac{(y_2- y_1)}{(x_2 - x_1)}(x - x_1)\\\\y - 2 = \frac{(0- 2)}{(4 -0)}(x -0)[/tex]

y - 2 = -1/2(x)

y = -1/2(x) + 2.

In this context, we can reasonably infer and logically deduce that an equation of the line that represents this graph in slope-intercept form is y = -1/2(x) + 2.

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there is an 85% chance that the two cards drawn at random from an ordinary deck of 52 cards will not be sixes.

The probability of drawing a card from an ordinary deck without replacement can be determined using the concept of conditional probability. Conditional probability is the probability of an event occurring, assuming that another event has already occurred.

In order to calculate the probability that the two cards drawn are not sixes, we can use the formula:

P(A and B) = P(A) x P(B|A)

Where A and B represent two independent events, P(A) is the probability of event A occurring, and P(B|A) is the conditional probability of event B occurring given that event A has already occurred.

The probability of drawing the first card that is not a six is:

P(A) = 48/52 = 0.9231

The probability of drawing the second card that is not a six, given that the first card drawn was not a six, is:

P(B|A) = 47/51 = 0.9216

Therefore, the probability of drawing two cards at random from an ordinary deck of 52 cards and having neither of them be a six is:

P(A and B) = P(A) x P(B|A) = 0.9231 x 0.9216 = 0.8503 or approximately 85%.

This means that there is an 85% chance that the two cards drawn at random from an ordinary deck of 52 cards will not be sixes.

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Answer: hope its help

Let's start by assigning variables to the unknown quantities in the problem. Let h be the height of the triangle in inches, and let b be the base of the triangle in inches.

According to the problem, the base of the triangle is 3 inches shorter than its height. This can be expressed as:

b = h - 3

The formula for the area of a triangle is:

A = (1/2)bh

We are given that the area of the triangle is 275 square inches, so we can substitute these values into the formula to get:

275 = (1/2)(h)(h-3)

Simplifying the right-hand side, we get:

275 = (1/2)(h^2 - 3h)

Multiplying both sides by 2 to eliminate the fraction, we get:

550 = h^2 - 3h

Rearranging this equation to standard quadratic form, we get:

h^2 - 3h - 550 = 0

Now we can solve for h using the quadratic formula:

h = (-b ± sqrt(b^2 - 4ac)) / (2a)

In this case, a = 1, b = -3, and c = -550, so we can substitute these values into the formula to get:

h = (-(-3) ± sqrt((-3)^2 - 4(1)(-550))) / (2(1))

Simplifying the expression inside the square root, we get:

h = (3 ± sqrt(2209)) / 2

We can ignore the negative solution since height must be positive, so we get:

h = (3 + sqrt(2209)) / 2 ≈ 29.04

Now that we know the height of the triangle is approximately 29.04 inches, we can use the equation b = h - 3 to find the length of the base:

b = 29.04 - 3 = 26.04

Therefore, the base of the triangle is approximately 26.04 inches, and the height is approximately 29.04 inches.

Step-by-step explanation:

Answer:

Yes. If you are asking if the duration between those two times is a total of 12 hours, the answer is yes.

Step-by-step explanation:

9:45am is 12 hours away from 9:45pm. This applies to all times and their am/pm counterparts such as 12am/12pm.