7 more than twice a number is 35.​

Answer:

105 degrees

Step-by-step explanation:

sum of angles in triangle is 180 degrees

11x-5+6x+5+x = 180

simplify this to get 18x=180

180/18 = 10 = x

plug in 10 for x

11(10) - 5

110-5

105

The possible expression that can be a rule for the number sequence is:[tex]$a_n = 4n + 1$[/tex],

As an illustration, the expression x + y is one where x and y are words with an addition operator in between. There are two types of expressions in mathematics: numerical expressions, which only comprise numbers, and algebraic expressions, which also include variables.

One possible expression that can be a rule for the number sequence is:

[tex]$a_n = 4n + 1$[/tex], where n is the position of the term in the sequence.

Using this expression, we can find the values of the first few terms as follows:

[tex]$a_1 = 4(1) + 1 = 5$[/tex]

[tex]$a_2 = 4(2) + 1 = 9$[/tex]

[tex]$a_3 = 4(3) + 1 = 13$[/tex]

[tex]$a_4 = 4(4) + 1 = 17$[/tex]

[tex]$a_5 = 4(5) + 1 = 21$[/tex]

Thus, possible expression that can be a rule for the number sequence is:[tex]$a_n = 4n + 1$[/tex],

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Answer:

5n, where n is equal to 0, 1, 2, 3, 4

Step-by-step explanation:

5, 9, 13, 17, 21,

The classification of student class designation (freshman, sophomore, junior, senior) is an example of a categorical random variable.  The correct option is A.

A random variable is a numerical or categorical quantity whose value is unknown but whose behavior can be forecast based on data that has been measured or observed. Random variables are typically used to represent quantities that fluctuate over time or are subject to chance occurrences.

The types of random variables are as follows:

i) Categorical random variable: This type of variable contains categorical data or data that are descriptive in nature. It is used to classify items or events into categories, which can be named or identified. For example, a set of data that includes categories like gender, eye color, or country of origin.

ii) Discrete random variable: This type of variable takes on discrete values, which means it can only take on whole numbers. For example, the number of cars sold at a dealership on any given day is a discrete random variable because it can only take on integer values.

iii) Continuous random variable: This type of variable takes on continuous values, which means it can take on any value within a given range. For example, the temperature in a room can take on any value between a certain minimum and maximum value.

Therefore, the correct option is A.

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The matrix A does not eigenvector v corresponding to the given eigen value.

λ.A = [ -11  60 ]

v = [ 1   3λ ]

v is an eigenvector of A calculate corresponding eigenvalue,

A × v = λ × v

where A is the given matrix.

v is the given vector.

λ is the corresponding eigenvalue.

× denotes matrix multiplication.

Let's first calculate A × v we have,

A × v

= [-11 60] × [ 1 3λ]

= [-11-33λ    60+ 180λ]

Check A× v is equal to λ × v ,

λ × v =

λ × [ 1 3λ]

= [λ  3λ^2]

Set these two vectors equal to each other and get the following system of equations ,

-11-33λ = λ  __(1)

60+ 180λ = 3λ^2  ___(2)

From equation (1) we get,

⇒34λ = -11

⇒ λ = -11/34

Substituting this value of λ into the second equation, we have,

60+ 180λ = 3λ^2

⇒ 60 + 180(-11/34) = 3(-11/34)^2

⇒ (2040 -1980)/ 34 = 3(-11/34)^2

⇒ 60/34 = 3( 11/34)^2

⇒ 60 × 34 = 3 × 11 × 11

⇒20× 34 = 11 × 11

Which is not true.

so the value of λ does not satisfies both equations.

Therefore, v is not an eigenvector of A with corresponding eigenvalue λ.

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The above question is incomplete, the complete question is:

Check whether v is an eigen vector of A  if yes find the corresponding eigen value.

λ.A = [ -11  60 ]

v = [ 1   3λ ]

Answer:I=(PxRxT)/100

I=(10000x20x1)/100x2

I=200000/200

I=1000

Step-by-step explanation:

Answer:

We can use the formula for compound interest to solve the problem:

A = P(1 + r/n)^(nt)

where A is the final amount, P is the principal, r is the annual interest rate, n is the number of times interest is compounded per year, and t is the number of years.

In this case, we know that P = $2,000, A = $6,883.55, n = 4 (quarterly compounding), and t = 16. We can solve for r by rearranging the formula as follows:

r = n[(A/P)^(1/nt) - 1]

Substituting the values, we get:

r = 4[(6,883.55/2,000)^(1/(4*16)) - 1] = 0.0522 or 5.22%

Therefore, the annual interest rate is approximately 5.22%

The length of the third side of the actual patio is 15.75 feet.

In order to solve issues requiring proportional connections between the sides of two triangles, it is crucial to understand the idea of triangle similarity. If the matching sides and angles of two triangles are identical, then two triangles are said to be similar. When referring to similarities, the symbol is used.

Let us suppose the length of third side = x.

Given that, blueprint of a triangular patio has side lengths 3.5 in., 3.5 in., and 5.25 in.

Thus, the blueprint and the actual patio form two similar triangles.

The sides of the similar triangles are equal in ratio thus,

3.5/5.25 = 10.5/x

3.5x = 5.25 * 10.5

x = (5.25 * 10.5) / 3.5

x = 15.75

Hence, the length of the third side of the actual patio is 15.75 feet.

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The area of the quadrilateral is 161.24 cm².

We can see that we can divide the quadrilateral into two triangles: ABD and CBD. We know that the height of ABD is 6 cm and the height of CBD is 8 cm. We also know that BD is 15 cm. To find the area of each triangle, we need to find the base of each triangle. We can do this using the Pythagorean theorem.

For triangle ABD:

AB² = AD² + BD²

AB² = (6 cm)² + (15 cm)²

AB² = 261 cm²

AB = [tex]\sqrt(261) cm[/tex]

For triangle CBD:

BC² = CD² + BD²

BC² = (8 cm)² + (15 cm)²

BC² = 289 cm²

BC = 17 cm

Now we can find the areas of the triangles:

Area of ABD =[tex]\frac{1}{2}[/tex] * AB * 6 cm

Area of ABD = [tex]\frac{1}{2}[/tex] * [tex]\sqrt(261) cm[/tex] * 6 cm

Area of ABD = 93.24 cm^2

Area of CBD = [tex]\frac{1}{2}[/tex] * BC * 8 cm

Area of CBD = [tex]\frac{1}{2}[/tex] * 17 cm * 8 cm

Area of CBD = 68 cm²

Finally, we can find the area of the quadrilateral by adding the areas of the triangles:

Area of ABCD = Area of ABD + Area of CBD

Area of ABCD = 93.24 cm² + 68 cm²

Area of ABCD = 161.24 cm²

Therefore, the area of the quadrilateral is 161.24 cm².

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Answer:

Tank 1: 6 litres

Tank 2: 6.75 litres

Tank 3: 4.5 litres

Step-by-step explanation:

Initially, James had 18 litres of water, and he poured three-quarters of it from tank 1 into tank 2. This means that the volume of water left in tank 1 is:

18 - (3/4) * 18 = 4.5 litres

The volume of water in tank 2 is:

(3/4) * 18 = 13.5 litres

Next, he poured half of the water in tank 2 (which is now 13.5 litres) into tank 3. The volume of water left in tank 2 is:

(1/2) * 13.5 = 6.75 litres

The volume of water in tank 3 is:

13.5 * (1/2) = 6.75 litres

Finally, he poured one-third of the water in tank 3 (which is now 6.75 litres) into tank 1. The volume of water in tank 1 after this is:

4.5 + (1/3) * 6.75 = 6 litres

The volume of water in tank 3 after this is:

6.75 * (2/3) = 4.5 litres

So the final volume of water in each tank is:

Tank 1: 6 litres

Tank 2: 6.75 litres

Tank 3: 4.5 litres

Based on the following sorted 20 values for age, the possible split points are {20, 21, 23, 25, 27, 29. 5, 31. 5, 32. 5, 34, 37. 5, 41, 42. 5, 44, 46, 48, 49. 5, 51, 52, 54, 56} (option a).

Option A suggests that the split points are {20, 21, 23, 25, 27, 29.5, 31.5, 32.5, 34, 37.5, 41, 42.5, 44, 46, 48, 49.5, 51, 52, 54, 56}. Notice that every split point falls between two consecutive ages in the original list. For example, the first split point is 20 because it is between 20 and 22. The second split point is 21 because it is between 20 and 22 as well.

Option B suggests that the split points are {21, 23, 25, 27, 29.5, 31.5, 32.5, 34, 37.5, 41, 42.5, 44, 46, 48, 49, 51, 52.5, 54, 56, 57}. Notice that the only difference between this option and Option A is that the last split point is 57 instead of 49.5.

Option C suggests that the split points are {0, 21, 23, 25, 27, 29.5, 31.5, 32.5, 34, 37.5, 41, 42.5, 44, 46, 48, 49, 51, 52.5, 54, 56}. Notice that the first split point is 0, which is not a possible age in the original list.

Option D suggests that the split points are {21, 23, 25, 27, 29.5, 31.5, 32.5, 34, 37.5, 41, 42.5, 44, 46, 48, 49.5, 51, 52.5, 54, 56}. Notice that the only difference between this option and Option A is that the split point after 49 is 49.5 instead of 49.5.

In summary, the correct answer is Option A because it provides all the possible split points that fall between the ages in the original list. When working with split points, it's important to consider the specific context and criteria for dividing the data.

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Answer:

n(s) =52. n(f) = 12 n(h) = 13

p (f)= 13/52. p(f)= 12/52

The expression's value is when x 0, and since |x| = -x when x 0, x + |x| simplifies to 0. In this case, x + |x| = x + (-x) = 0 for x 0.

When the variables and constants in a mathematical expression are given values, the outcome of the computation it describes is the expression's value. The value of a function, given the value(s) assigned to its argument, is the sum that the function assumes for these input values (s).

For x =7,x+|x| =7+|7| =14

For x =10,x+|x|= 10+|10| =20

For x = 0,x+|x| =0+|0| =0

For x = -3, x + |x| = -3 + |-3| = 0

For x = -8, x + |x| = -8 + |-8| = 0

The expression's value is when x 0, and since |x| = -x when x 0, x + |x| simplifies to 0. In this case, x + |x| = x + (-x) = 0 for x 0.

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3. The statement is true,  the correlation coefficient is close to -1. 4. temperature for a city with a latitude of 48 is 43. 5. The statement is false. 6. cannot make a reasonable estimate

Equations can be used to model real-world situations and solve problems in many fields, including science, engineering, finance, and more. They are an essential tool in mathematics and are used extensively in algebra, calculus, and other advanced branches of math.

Equations can involve various mathematical operations such as addition, subtraction, multiplication, division, exponentiation, and others.

Question 3:

The statement is true. We can check this by calculating the correlation coefficient between the latitude and temperature data points, which should be close to -1. The calculated line of best fit is also consistent with the given data.

Question 4:

To estimate the temperature for a city with a latitude of 48, we can use the equation of the line of best fit:

y = -1.07x + 92.87

Substituting x = 48, we get:

y = -1.07(48) + 92.87

y = 42.79

Rounding to the nearest whole number, the estimated temperature for a city with a latitude of 48 is 43.

Question 5:

The statement is false. We can check this by calculating the correlation coefficient between the passengers and suitcases data points, which should be close to 1. The given line of best fit has a negative slope, which is inconsistent with the positive correlation between the variables.

Question 6:

To estimate the number of suitcases for a flight carrying 250 people, we can use the equation of the line of best fit:

y = -1.98x + 7.97

Substituting x = 250, we get:

y = -1.98(250) + 7.97

y = -485.03

However, it does not make sense for the number of suitcases to be negative. Therefore, we cannot make a reasonable estimate for the number of suitcases on a flight carrying 250 people using this line of best fit.

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By answering the presented question, we may conclude that I used the following procedures to produce this graph.

Mathematicians use graphs to visually display or chart facts or values in order to express them coherently. A graph point usually represents a connection between two or more items. A graph, a non-linear data structure, is made up of nodes (or vertices) and edges. Glue the nodes, also known as vertices, together. This graph contains vertices V=1, 2, 3, 5, and edges E=1, 2, 1, 3, 2, 4, and (2.5), (3.5). (4.5). Statistical graphs (bar graphs, pie graphs, line graphs, and so on) are graphical representations of exponential development. a logarithmic graph shaped like a triangle.

I used the following procedures to produce this graph:

I classified the personnel as full-time, part-time, and temporary.

I estimated the proportion of employees who assessed the company's work-life balance as "very good" or "excellent" for each employee category, as well as the percentage who rated it as "good" or "fair/poor."

I used the following procedures to produce this graph:

I classified the personnel as full-time, part-time, and temporary.

I estimated the proportion of employees who assessed the company's work-life balance as "very good" or "excellent" for each employee category, as well as the percentage who rated it as "good" or "fair/poor."

I made the segmented bar graph using these percentages.

The graph was made using Excel technology. You may make a similar graph with Excel or any other software that supports segmented bar graphs.

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The set of points that represent the intersection of the curve of vector function F and circle C is

{(4+10cos(t), 4+10cos(t), 16+40cos(t)) | t ranges from 0 to 2π}.

We have,

Vector function F = (y + 2x, 2x + 5z, 7y + 8x)

C is the circle with radius 5, center at (2,0,0), in the plane x = 2

C is oriented counterclockwise as viewed from the origin (0,0,0)

The vector function F represents a three-dimensional curve in space.

The circle C is a two-dimensional object in space, lying in the plane x = 2 and centered at (2,0,0) with a radius of 5. It is also oriented counterclockwise as viewed from the origin (0,0,0).

To find the intersection of vector function curve F and circle C, we can substitute the equation of the circle into the equation of the curve and solve for the parameter(s) that satisfy the equation. However, since the equation of the circle is given in terms of x only, we can simplify the equation of the curve by substituting y = 0 and z = 0:

F = (2x, 2x, 8x)

Now, we can substitute x = 2 + 5cos(t) and y = 5sin(t) (the parameterization of the circle C in the plane x = 2) into the equation of the curve F:

F = (2(2+5cos(t)), 2(2+5cos(t)), 8(2+5cos(t)))

= (4+10cos(t), 4+10cos(t), 16+40cos(t))

Thus, the intersection of vector function curve F and circle C is given by the set of points:

{(4+10cos(t), 4+10cos(t), 16+40cos(t)) | t in [0, 2π)}

Note- that the parameter t represents the angle of rotation around circle C, and ranges from 0 to 2π to cover the entire circle.

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P(∅) = 0, P(A∪B) = 0.4 , P(A∩B) = 0.1 ,Since the sample space is not defined in the question, we cannot calculate P(B'). Therefore, we cannot calculate P(A∩B').and  P(A∪B) = 0.4. are the required solutions ofgiven probability check .

a. The probability of an empty set is always zero. Therefore, P(∅) = 0.

b. The probability of the union of two events, A and B, is given by the formula P(A∪B) = P(A) + P(B) - P(A∩B). Substituting the values given in the question, we get:

P(A∪B) = P(A) + P(B) - P(A∩B)

= 0.3 + 0.2 - 0.1

= 0.4

Therefore, P(A∪B) = 0.4.

c. The probability of the intersection of A and B is given by the formula P(A∩B). Substituting the values given in the question, we get:

P(A∩B) = 0.1

Therefore, P(A∩B) = 0.1.

d. The probability of the intersection of A and the complement of B is given by the formula P(A∩B'). The complement of B is the set of all outcomes that are not in B. Since the sample space is not defined in the question, we cannot calculate P(B'). Therefore, we cannot calculate P(A∩B').

e. The probability of the union of A and B is given by the formula P(A∪B). Substituting the values given in the question, we get:

P(A∪B) = P(A) + P(B) - P(A∩B)

= 0.3 + 0.2 - 0.1

= 0.4

Therefore, P(A∪B) = 0.4.

In probability theory, the union of two events A and B is the set of outcomes that belong to either A or B or both. The intersection of two events A and B is the set of outcomes that belong to both A and B. The complement of an event A is the set of outcomes that do not belong to A. These concepts are fundamental in probability theory and are used extensively in solving various problems.

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New points of graph A'B'C'D' are A'(-2, -2), B'(-2, 0), C'(-4, 0), D'(-4, -1)

In graph theory, the term "translation" refers to a type of operation that moves all the vertices and edges of a graph by a fixed distance in a given direction. Specifically, a translation of a graph involves shifting every vertex a certain distance horizontally and/or vertically, without changing the shape or connectivity of the graph.

Translation: 4 left and 2 down

Start with a point at its original location and then move it 4 units to the left and 2 units down. This can be done by subtracting 4 from the x-coordinate and subtracting 2 from the y-coordinate of the point or shape.

Given points in a graph ABCD are, A(2, 0), B(2, 2), C(0, 2), D(0, 1)

Subtract 4 from the x-coordinate and subtract 2 from the y-coordinate, resulting in a new points of graph A'B'C'D' are A'(-2, -2), B'(-2, 0), C'(-4, 0), D'(-4, -1)

The figure shown in below diagram.

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Therefore, with a 95% confidence interval, we can estimate the true mean length of the drive shafts to be between 992.16 mm and 1007.84 mm.

The true mean length of the drive shafts can be estimated using a 95% confidence interval. The margin of error is calculated as 2*1.96*2 = 7.84 mm. The lower bound of the confidence interval is 1000 - 7.84 = 992.16 mm and the upper bound is 1000 + 7.84 = 1007.84 mm.

This confidence interval states that there is a 95% probability that the true mean length of the drive shafts falls within the range of 992.16 mm to 1007.84 mm. The relationship between the two probabilities is that the probability of the true mean length falling within the confidence interval is 95%. If the sample size was increased, the margin of error would decrease, resulting in a tighter range and higher probability that the true mean would fall within the confidence interval.
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Let a sample space be partitioned into three mutually exclusive and exhaustive events, B1, B2, and, B3.We have to complete the following probability table, prior probabilities, conditional probabilities, joint probabilities, and posterior probabilities, and also we have to round our answers to 2 decimal places.

Given information:Probability of B1, P(B1) = 0.11 Probability of A given B1, P(A | B1) = 0.45 Probability of A intersection B1, P(A ∩ B1) = ?

Probability of B1 given A, P(B1 | A) = ?     Probability of B2, P(B2) = ?     Probability of A given B2, P(A | B2) = 0.62, Probability of A intersection B2,  P(A ∩ B2) = ?     Probability of B2 given A, P(B2 | A) = ?

Probability of B3, P(B3) = 0.38.    Probability of A given B3, P(A | B3) = 0.85.   Probability of A intersection B3,

P(A ∩ B3) = ?

Probability of B3 given A, P(B3 | A) = ?

Total probability of A, P(A) = ?

Total probability of sample space = 1.     Let's complete the probability table:Prior probabilities, Conditional probabilities, Joint probabilities, Posterior probabilities      P(B1) = 0.11P(A | B1) = 0.45P(A ∩ B1) = P(A | B1) * P(B1) / P(A)P(B1 | A) = P(A | B1) * P(B1) / P(A)P(B2) = 1 - (P(B1) + P(B3)) = 1 - (0.11 + 0.38) = 0.51P(A | B2) = 0.62P(A ∩ B2) = P(A | B2) * P(B2) / P(A)P(B2 | A) = P(A | B2) * P(B2) / P(A)P(B3) = 0.38P(A | B3) = 0.85P(A ∩ B3) = P(A | B3) * P(B3) / P(A)P(B3 | A) = P(A | B3) * P(B3) / P(A)Total = 1P(A) = P(A ∩ B1) + P(A ∩ B2) + P(A ∩ B3) = 0.11 * 0.45 + 0.51 * 0.62 + 0.38 * 0.85 = 0.6579.  So, the probability table is as follows:Prior probabilities,Conditional probabilities,Joint probabilities,Posterior probabilities

P(B1) = 0.11P(A | B1) = 0.45P(A ∩ B1) = 0.0495P(B1 | A) = 0.3419P(B2) = 0.51P(A | B2) = 0.62P(A ∩ B2) = 0.3162P(B2 | A) = 0.4857P(B3) = 0.38P(A | B3) = 0.85P(A ∩ B3) = 0.3217P(B3 | A) = 0.1724Total = 1P(A) = 0.6579

Hence, the completed probability table is as shown above.

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Answer:

Size 3 ‍♀️

Step-by-step explanation:

In the US, the average shoe size for 10-Year-Old is USA Size 3.

-Jul 12, 2020

The area is changing at a rate of 28,160 m²/year at that point in time.

The area of the rectangular region is given by:

A = lw

Where l is the length of the rectangular region and w is the width of the rectangular region.

The width of the rectangular region is given to be 220 m. Therefore, we have the width w = 220 m. The length l of the rectangular region can be found knowing that it is twice as long as it is wide. Therefore, the length of the rectangular region is given by:

l = 2w

l = 2 x 220

l = 440

Therefore, the length l of the rectangular region is 440 m.

At the given point in time, the width of the rectangular region is growing at a rate of 32 m per year. Therefore, we have the rate of change of the width dw/dt to be 32 m per year. We need to find how fast the area of the rectangular region is changing at that point in time. Therefore, we need to find the rate of change of the area of the rectangular region dA/dt.

A = lw

dA/dt = w dl/dt + l dw/dt

dA/dt = 220 d/dt(2w) + 440 dw/dt

dA/dt = 220 x 2 dw/dt + 440 dw/dt

dA/dt = 880 dw/dt

Substitute the value of dw/dt to get:

dA/dt = 880 x 32

dA/dt = 28,160 m²/year

Therefore, the area of the rectangular region has a rate of change of 28,160 m² per year at that point in time.

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The scientific and standard notation and clue obtained from the clue sheet are;

1. Name starts with; J

2. Difference = 145 million miles = Weight = 145 lbs

3. Height: 5 ft, 6 in

4. I. 2.54 × 10⁸ miles corresponding to letter I

II. 0.0005825 corresponds to letter G

III. 5.0432 × 10⁶ corresponds to letter N

IV. 1.547 × 10³ corresponds to letter R

V. 4.977 × 10⁻² corresponds to letter W

VI. 1.3 × 10¹⁰ light years away; corresponds to letter D

VII. 5.04 × 10⁻⁵ corresponds to letter A

The suspects hubby is DRAWING

Scientific notation is a format used to express very large or very small numbers such that they are much easier to work with. The scientific notation format is; a × 10ⁿ, where; a is the coefficient, which is a number between 1 and 10 (10 excluded), and n is an integer.

1. G = (6.07 × 10⁷)/(7.035 × 10³) ≈ 8628.29

J = (6.03 × 10⁻³)/(5.05 × 10⁻⁷) ≈ 11940.59

Therefore; J > G

The suspects name starts with J

2. The distance the telescope in the laboratory allows the viewer to see = 1.5 × 10⁹ miles away

The distance the other telescope a few hours away allows the viewer to see = 1.355 × 10⁹ miles away

The difference between the distances = (1.5 - 1.355) × 10⁹ miles = 1.45 × 10⁸ miles

The difference in the distance is 1.45 × 10⁸ miles = 145 million miles

The suspect weight is 145 lbs

3. The numbers are;

Feet; 532.063 × 10³ = 5.32063 × 10⁵

Inches; 5,030,045 = 5.030045 × 10⁶

The height of the suspect is 5 feet 6 inches (5'6'') = 5.5 feet

Height; = 5 ft, 6 in

4. I. The difference in distances between Earth and Saturn can be found as follows;

The difference in the distances = (1000 - 746) million miles = 254 million miles apart

Scientific notation is the expression of numbers in the form consisting of a number between 1 and 10, multiplied by 10 raised to a power

254 million miles = 2.54 × 10⁸ miles

The corresponding letter from the code cracker is; I

II Standard notation is the expression of numbers in the standard form without the use of exponents or special symbols

The number 5.825 × 10⁻⁴ in standard notation is; 0.0005825

The corresponding letter from the code cracker is; G

III. The number 504.32 × 10⁴ in scientific notation can be obtained by moving the decimal point two places to the left followed by increasing the index of 10 by 2 as follows;

504.32 × 10⁴ = 5.0432 × 10⁶

The corresponding letter from the code cracker is; N

IV. The sum of the numbers 1.202 × 10³ and 3.45 × 10² can be obtained by expressing both numbers to the same power of 10 as follows;

1.202 × 10³ + 3.45 × 10² = 12.02 × 10² + 3.45 × 10² = 15.47 × 10²

15.47 × 10² = 1.547 × 10³

Therefore; 1.202 × 10³ + 3.45 × 10² = 1.547 × 10³

The corresponding letter from the code cracker is; R

V. The difference of the numbers can be obtained as follows;

5.023 × 10⁻² - 4.6 × 10⁻⁴ = 502.3 × 10⁻⁴ - 4.6 × 10⁻⁴ = 497.7 × 10⁻⁴

497.7 × 10⁻⁴ = 4.977 × 10⁻²

Therefore; 5.023 × 10⁻² - 4.6 × 10⁻⁴ = 4.977 × 10⁻²

The corresponding letter from the code cracker is; W

VI. 13 billion light years = 13 × 10⁹ light years = 1.3 × 10¹⁰ light years

The distance a standard telescope can allow to be seen is 1.3 × 10¹⁰ light years away

The corresponding letter from the code cracker is; D

VII. 0.0000504 in scientific notation is; 5.04 × 10⁻⁵

The corresponding letter from the code cracker is; A

IGNRWDA

The suspects favorite hubby is DRAWING

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Julie will have driven a total distance of 289 miles if she travels from York to Corby via Derby.

Starting from York, Julie needs to travel to Derby. The distance between York and Derby is given as 89 miles. So, we know that Julie will have driven 89 miles once she reaches Derby.

Next, Julie needs to travel from Derby to Corby, but the given information is a bit tricky here. The distance from Derby to Corby is not given directly. Instead, we are given two distances - Derby to Dory and Dory to Corby.

To find the distance from Derby to Corby, we need to add the distances between Derby and Dory, and Dory and Corby. From the question, we know that the distance between Derby and Dory is 127 miles and the distance between Dory and Corby is 73 miles. Adding these two distances gives us the total distance from Derby to Corby, which is 200 miles.

Finally, we can add up the distances traveled between each location to find the total distance traveled by Julie. Adding the distances of each leg of the journey, we get:

89 miles (York to Derby) + 200 miles (Derby to Corby via Dory) = 289 miles

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Complete Question:

If Julie drives from York to Corby via Dory how many miles will she have driven?

York 89

Derby 127 73

Corby

The length οf line FG is 12 cm, If  Figure EFGH is a parallelοgram.

A parallelοgram is a type οf quadrilateral with twο pairs οf parallel sides. The οppοsite sides οf a parallelοgram are equal in length and parallel tο each οther.

Since EFGH is a parallelοgram, we knοw that the οppοsite sides are parallel and equal in length. Therefοre, the length οf line FG is equal tο the length οf line EH.

We can find the length οf EH by using the Pythagοrean theοrem οn right triangle EFG:

[tex]EF^2 + FG^2 = EG^2[/tex]

Since EF = 5 cm, EG = 13 cm, and angle FEG is a right angle (as οppοsite angles in a parallelοgram are equal), we can sοlve fοr FG:

[tex]FG^2 = EG^2 - EF^2[/tex]

[tex]FG^2 = 13^2 - 5^2[/tex]

[tex]FG^2 = 144[/tex]

[tex]FG = \sqrt{(144)[/tex]

[tex]FG = 12 cm[/tex]

Therefοre, the length οf line FG is 12 cm.

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The equation of the line perpendicular to y = 1/3x - 9 that passes through the point (-6, 10) is y = -3x - 8.

The given equation is in slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept.

So, we can see that the slope of the given line is 1/3.

A line perpendicular to this line will have a slope that is the negative reciprocal of the slope of the given line.

The negative reciprocal of 1/3 is -3.

Now, we have the slope of the perpendicular line and a point that it passes through. We can use point-slope form to find the equation of the line.

Point-slope form, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope.

Substituting the values we have

y - 10 = -3(x - (-6))

y - 10 = -3(x + 6)

y - 10 = -3x - 18

y = -3x - 8

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The given question is incomplete, the complete question is:

Write the equation of a line perpendicular to y = 1/3x - 9 that passes through the point (-6, 10)

using the ratio given, the number of computers could be sold by the company last week is: A. 448 desktops, 168 laptops.

To find the actual numbers of desktop and laptop computers sold, we need to choose a common factor for the ratio 8:3.

Let's assume that the total number of computers sold is 33x (where x is a positive integer). Then, the ratio 8:3 corresponds to 8x desktops and 3x laptops. We can check which of the given options satisfies this condition:

A. 8x = 448, 3x = 168 --> This satisfies the condition, as 8:3 = 448:168

B. 8x = 448, 3x = 165 --> This does not satisfy the condition, as 8:3 is not equal to 448:165

C. 8x = 440, 3x = 168 --> This does not satisfy the condition, as 8:3 is not equal to 440:168

D. 8x = 400, 3x = 165 --> This does not satisfy the condition, as 8:3 is not equal to 400:165

Therefore, the answer is option A: 448 desktops and 168 laptops could be the numbers of computers sold by the company last week.

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Answer: If you end up with a remainder of 0, then you have a terminating decimal. Otherwise, the remainders will begin to repeat after some point, and you have a repeating decimal.

Step-by-step explanation:

I hope that this helped! :)

Given:

5× ______ = - 35

• -35/5

• -7

5 x -7 = -35

Answer:

The answer is 7

Step-by-step explanation:

Divide each term in 5x=-35 by 5 and simplify.x=−7

a. Hence proved that the sum of fractions  [tex]${\frac{1}{\sqrt{1+\sqrt{2}}}}+{\frac{1}{\sqrt{2+\sqrt{3}}}}+{\frac{1}{\sqrt{3}+\sqrt{4}}}=1$[/tex]

b. The value will be 7 for the expression

⇒ [tex]${\frac{ 1-\sqrt{2}}{-1}+{\frac{ \sqrt{2}-\sqrt{3}}{-1}+ \cdot \cdot \cdot+{\frac{\sqrt{63}-8}{-1}}$[/tex]

Square rοοt οf a number is a value, which οn multiplicatiοn by itself, gives the οriginal number. The square rοοt is an inverse methοd οf squaring a number. Hence, squares and square rοοts are related cοncepts.

Suppοse x is the square rοοt οf y, then it is represented as x=√y, οr we can express the same equatiοn as x² = y. Here, ‘√’ is the radical symbοl used tο represent the rοοt οf numbers. The pοsitive number, when multiplied by itself, represents the square οf the number. The square rοοt οf the square οf a pοsitive number gives the οriginal number.

Here,

a. [tex]${\frac{1}{\sqrt{1+\sqrt{2}}}}+{\frac{1}{\sqrt{2+\sqrt{3}}}}+{\frac{1}{\sqrt{3}+\sqrt{4}}}=1$[/tex]

Using (a + b)(a - b) = a² - b²

⇒ [tex]${\frac{1 \cdot \sqrt{1}-\sqrt{2}}{\sqrt{1}+\sqrt{2}\cdot \sqrt{1 }-\sqrt{2}}+{\frac{1 \cdot \sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}\cdot \sqrt{1}-\sqrt{2}}}+{\frac{1 \cdot \sqrt{3}-\sqrt{4}}{\sqrt{3}+\sqrt{4}\cdot \sqrt{3}-\sqrt{4}}}$[/tex]

⇒ [tex]${\frac{ \sqrt{1}-\sqrt{2}}{1-2}+{\frac{ \sqrt{2}-\sqrt{3}}{2-3}+{\frac{\sqrt{3}-\sqrt{4}}{3-4}}$[/tex]

⇒ [tex]${\frac{ \sqrt{1}-\sqrt{2}}{-1}+{\frac{ \sqrt{2}-\sqrt{3}}{-1}+{\frac{\sqrt{3}-\sqrt{4}}{-1}}$[/tex]

⇒ [tex]${\frac{ 1-\sqrt{2}}{-1}+{\frac{ \sqrt{2}-\sqrt{3}}{-1}+{\frac{\sqrt{3}-2}{-1}}$[/tex]

⇒ [tex]${\frac{ 1-\sqrt{2}}{-1}+{\frac{ \sqrt{2}-\sqrt{3}}{-1}+{\frac{\sqrt{3}-2}{-1}}$[/tex]

⇒ [tex]$ -1+\sqrt{2}}- \sqrt{2}+\sqrt{3}}-{\sqrt{3}+2}$[/tex]

⇒ [tex]$ -1+2}$[/tex]

⇒ 1

a. Hence proved that the sum of fractions  [tex]${\frac{1}{\sqrt{1+\sqrt{2}}}}+{\frac{1}{\sqrt{2+\sqrt{3}}}}+{\frac{1}{\sqrt{3}+\sqrt{4}}}=1$[/tex]

B. This will be done with the same process,

⇒ [tex]${\frac{ 1-\sqrt{2}}{-1}+{\frac{ \sqrt{2}-\sqrt{3}}{-1}+ \cdot \cdot \cdot+{\frac{\sqrt{63}-8}{-1}}$[/tex]

⇒ [tex]${\frac{ 1-\sqrt{2}}{-1}+{\frac{ \sqrt{2}-\sqrt{3}}{-1}+ \cdot \cdot \cdot+{\frac{\sqrt{63}-8}{-1}}$[/tex]

⇒ [tex]$ -1+\sqrt{2}}- \sqrt{2}+\sqrt{3}} \cdot \cdot \cdot -{\sqrt{63}+8}$[/tex]

There, will be same roots of every number until - 8

So,

⇒ [tex]$ -1+8}$[/tex]

= 7

b. The value will be 7 for the expression

⇒ [tex]${\frac{ 1-\sqrt{2}}{-1}+{\frac{ \sqrt{2}-\sqrt{3}}{-1}+ \cdot \cdot \cdot+{\frac{\sqrt{63}-8}{-1}}$[/tex]

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