Find the area of the shaded region. Round to the nearest tenth. 11.1m 130°
Area = [ ? ] m²​

Given:

The graph of a line.

To find:

The equation for the given line.

Solution:

From the given graph, it is clear that the line passes through the points (0,-5) and (5,0). So, the equation of the line is:

[tex]y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)[/tex]

[tex]y-(-5)=\dfrac{0-(-5)}{5-0}(x-0)[/tex]

[tex]y+5=\dfrac{5}{5}(x)[/tex]

[tex]y+5=x[/tex]

Subtract 5 from both sides.

[tex]y+5-5=x-5[/tex]

[tex]y=x-5[/tex]

Therefore, the equation of the given line is [tex]y=x-5[/tex].

Answer:

[tex]y=-\frac{3}{2}x+1[/tex]

Step-by-step explanation:

Hi there!

What we need to know:

1) Determine the slope (m)

[tex]3x+2y = 10[/tex]

First, we must organize this given equation in slope-intercept form. This will help us identify its slope.

[tex]3x+2y = 10[/tex]

Subtract 3x from both sides

[tex]2y = -3x+10[/tex]

Divide both sides by 2

[tex]y = -\frac{3}{2} x+5[/tex]

Now, we can identify clearly that [tex]-\frac{3}{2}[/tex] is in the place of m in [tex]y=mx+b[/tex], making it the slope. Because parallel lines have the same slope, this makes the slope of the line we're currently solving for [tex]-\frac{3}{2}[/tex] as well. Plug this number into [tex]y=mx+b[/tex]:

[tex]y=-\frac{3}{2}x+b[/tex]

2) Determine the y-intercept (b)

[tex]y=-\frac{3}{2}x+b[/tex]

Plug in the given point (8,-11) and solve for b

[tex]-11=-\frac{3}{2}(8)+b\\-11=-\frac{24}{2}+b\\-11=-12+b[/tex]

Add 12 to both sides

[tex]1=b[/tex]

Therefore, the y-intercept of the line is 1. Plug this back into [tex]y=-\frac{3}{2}x+b[/tex]:

[tex]y=-\frac{3}{2}x+1[/tex]

I hope this helps!

Answer:

12

Step-by-step explanation:

PEMDAS is the order

P = parenthesis

E = exponent

M = *

D = division

A = +

S = -

so first 8*2 = 16

and then -8/2 = -4

and then -4 + 16

= 12

Answer:

a. -3

Step-by-step explanation:

-2 turns into 6 by multiplying it by -3.

the same from 6 to -18, from -18 to 54, ...

so, an/an+1 = -3

Answer:

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 \sin^2y}{x^2+2y^2} = 0[/tex]

Step-by-step explanation:

Given

[tex]f(x,y) = \frac{x^2 \sin^2y}{x^2+2y^2}[/tex]

Required

[tex]\lim_{(x,y) \to (0,0)} f(x,y)[/tex]

[tex]\lim_{(x,y) \to (0,0)} f(x,y)[/tex] becomes

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 \sin^2y}{x^2+2y^2}[/tex]

Multiply by 1

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 \sin^2y}{x^2+2y^2}\cdot 1[/tex]

Express 1 as

[tex]\frac{y^2}{y^2} = 1[/tex]

So, the expression becomes:

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 \sin^2y}{x^2+2y^2} \cdot \frac{y^2}{y^2}[/tex]

Rewrite as:

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 y^2}{x^2+2y^2} \cdot \frac{\sin^2y}{y^2}[/tex]

In limits:

[tex]\lim_{(x,y) \to (0,0)} \frac{\sin^2y}{y^2} \to 1[/tex]

So, we have:

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 y^2}{x^2+2y^2} *1[/tex]

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 y^2}{x^2+2y^2}[/tex]

Convert to polar coordinates; such that:

[tex]x = r\cos\theta;\ \ y = r\sin\theta;[/tex]

So, we have:

[tex]\lim_{(x,y) \to (0,0)} \frac{(r\cos\theta)^2 (r\sin\theta;)^2}{(r\cos\theta)^2+2(r\sin\theta;)^2}[/tex]

Expand

[tex]\lim_{(x,y) \to (0,0)} \frac{r^4\cos^2\theta\sin^2\theta}{r^2\cos^2\theta+2r^2\sin^2\theta}[/tex]

Factor out [tex]r^2[/tex]

[tex]\lim_{(x,y) \to (0,0)} \frac{r^4\cos^2\theta\sin^2\theta}{r^2(\cos^2\theta+2\sin^2\theta)}[/tex]

Cancel out [tex]r^2[/tex]

[tex]\lim_{(x,y) \to (0,0)} \frac{r^2\cos^2\theta\sin^2\theta}{\cos^2\theta+2\sin^2\theta}[/tex]

[tex]\lim_{(x,y) \to (0,0)} \frac{r^2\cos^2\theta\sin^2\theta}{\cos^2\theta+2\sin^2\theta}[/tex]

Express [tex]2\sin^2 \theta[/tex] as [tex]\sin^2\theta+\sin^2\theta[/tex]

So:

[tex]\lim_{(x,y) \to (0,0)} \frac{r^2\cos^2\theta\sin^2\theta}{\cos^2\theta+\sin^2\theta+\sin^2\theta}[/tex]

In trigonometry:

[tex]\cos^2\theta + \sin^2\theta = 1[/tex]

So, we have:

[tex]\lim_{(x,y) \to (0,0)} \frac{r^2\cos^2\theta\sin^2\theta}{1+\sin^2\theta}[/tex]

Evaluate the limits by substituting 0 for r

[tex]\frac{0^2 \cdot \cos^2\theta\sin^2\theta}{1+\sin^2\theta}[/tex]

[tex]\frac{0 \cdot \cos^2\theta\sin^2\theta}{1+\sin^2\theta}[/tex]

[tex]\frac{0}{1+\sin^2\theta}[/tex]

Since the denominator is non-zero; Then, the expression becomes 0 i.e.

[tex]\frac{0}{1+\sin^2\theta} = 0[/tex]

So,

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 \sin^2y}{x^2+2y^2} = 0[/tex]

Answer:

Buy 2 lollipops for $2.

Step-by-step explanation:

If you divide the total price by total items purchased, you get the price per unit. 2/2=1 or around $1, while 40/30=4/3 or around $1.3. You are paying 1$ per lollipop for the 2 lollipop choice and paying 1.3 dollars per lollipop for the 30 lollipop choice.

Answer:

3x+4

Step-by-step explanation:

When you factor 9x^2+24x+16, it factors to (3x+4)^2

Factoring 9x^2 - 16 factors to (3x+4)(3x-4)

Therefore the common factor is 3x+4

I hope this helps!

9514 1404 393

Answer:

Step-by-step explanation:

Of the items sold, sodas were 2/(2+1) = 2/3 of the total.

  (2/3)(228) = 152 . . . sodas were sold

  152/2 = 76 . . . . hot dogs were sold

Answer:

Step-by-step explanation:

Since, CD is an altitude, ∠CDB will be a right angle.

m∠CDB = m∠CDA = 90°

By applying triangle sum theorem in ΔABC,

m∠CAB + m∠CBA + m∠ACB = 180°

20° + m∠CBA + 90° = 180°

m∠CBA = 180° - 110°

             = 70°

Therefore, m∠CBD = 70°

By applying triangle sum theorem in ΔBCD,

m∠BCD + m∠CDB + m∠DBC = 180°

m∠BCD + 90° + 70° = 180°

m∠BCD + 160° = 180°

m∠BCD = 20°

m∠CAD = m∠A = 20°

m∠ACD = 90° - m∠BCD

             = 90° - 20°

m∠ACD = 70°

Answer: [tex]7\sqrt{42}[/tex]

Step-by-step explanation:

42 students divided equally into 7 groups means 42 divided by 7, and [tex]7\sqrt{42}[/tex] is the only choice that shows that.

7√42. is the expressions would represent a class of 42 students divided equally into 7 groups

A division is a process of splitting a specific amount into equal parts.

Given,

A class of forty two Forty two students divided equally into 7 groups.

Forty two students divided equally into seven groups means forty two divided by seven, and

this can be done by using 7√42.

42 students divided equally into 7 groups means forty two divided by seven

Hence  7√42. is the expressions would represent a class of 42 students divided equally into 7 groups

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Answer:

The angle for the other interior angel is 75°, all you have to do is subtract 180, from the 105

Step-by-step explanation:

Answer:

The system is inconsistent

Step-by-step explanation:

Given the matrix system solutions in the attachment, we can see that the coefficient of the matrix rank has become zero. This shows that it has no solution.

Answer:

1. x + 4 = 9

Hint: the word 'sum' generally refers to addition.

2. 10a = 70

3. [tex]\frac{3}{4} t[/tex] = 15

4. [tex]\frac{1}{4} x[/tex] - 4 = 4

the answer is c i just had this question your welcome

Answer:

C'

Step-by-step explanation:

Given

ABCD to A'B'C'D'

Required

Corresponding angle of C

ABCD to A'B'C'D' means that the following angles are corresponding

[tex]A \to A'[/tex]

[tex]B \to B'[/tex]

[tex]C \to C'[/tex]

[tex]D \to D'[/tex]

Hence, C' corresponds to C

Answer:

C

Step-by-step explanation:

I took the test :)

Answer:

TAMA PO

Step-by-step explanation:

Paki brainliest nadin po

Answer:

[tex]14x cos(\frac{1}{15}x^{2})=14 \sum _{k=0} ^{\infty} \frac{(-1)^{k}x^{4k+1}}{(2k)!15^{2k}}[/tex]

Step-by-step explanation:

In order to find this Maclaurin series, we can start by using a known Maclaurin series and modify it according to our function. A pretty regular Maclaurin series is the cos series, where:

[tex]cos(x)=\sum _{k=0} ^{\infty} \frac{(-1)^{k}x^{2k}}{(2k)!}[/tex]

So all we need to do is include the additional modifications to the series, for example, the angle of our current function is: [tex]\frac{1}{15}x^{2}[/tex] so for

[tex]cos(\frac{1}{15}x^{2})[/tex]

the modified series will look like this:

[tex]cos(\frac{1}{15}x^{2})=\sum _{k=0} ^{\infty} \frac{(-1)^{k}(\frac{1}{15}x^{2})^{2k}}{(2k)!}[/tex]

So we can use some algebra to simplify the series:

[tex]cos(\frac{1}{15}x^{2})=\sum _{k=0} ^{\infty} \frac{(-1)^{k}(\frac{1}{15^{2k}}x^{4k})}{(2k)!}[/tex]

which can be rewritten like this:

[tex]cos(\frac{1}{15}x^{2})=\sum _{k=0} ^{\infty} \frac{(-1)^{k}x^{4k}}{(2k)!15^{2k}}[/tex]

So finally, we can multiply a 14x to the series so we get:

[tex]14xcos(\frac{1}{15}x^{2})=14x\sum _{k=0} ^{\infty} \frac{(-1)^{k}x^{4k}}{(2k)!15^{2k}}[/tex]

We can input the x into the series by using power rules so we get:

[tex]14xcos(\frac{1}{15}x^{2})=14\sum _{k=0} ^{\infty} \frac{(-1)^{k}x^{4k+1}}{(2k)!15^{2k}}[/tex]

And that will be our answer.

Answer:

4.17%

(1/4)(1/3)(1/2)(1)

alternative you can say that there are 24 permutations of

4 items and that you have to guess 1 of them 1/24 = 4.17%

Step-by-step explanation:

0.25  

0.333333333  

0.5  

1  

Answer:

a = 0.0040 m/s², v = 14.4 m/s.

Step-by-step explanation:

Given that,

The distance between Kathmandu and Khanikhola, d = 26 km = 26000 m

Time, t = 1 hour = 3600 seconds

Let a is the acceleration of the bus. Using second equation of motion,

[tex]d=ut+\dfrac{1}{2}at^2[/tex]

Where

u is the initial speed of the bus, u = 0

So,

[tex]d=\dfrac{1}{2}at^2\\\\a=\dfrac{2d}{t^2}\\\\a=\dfrac{2\times 26000}{(3600)^2}\\\\a=0.0040\ m/s^2[/tex]

Now using first equation of motion.

Final velocity, v = u +at

So,

v = 0+0.0040(3600)

v = 14.4 m/s

Hence, this is the required solution.

Option D. The set of transformation that is performed on this triangle (ABCD)' is A translation 5 units to the right, followed by a 180-degree counterclockwise rotation about the origin

The answer choices have been shown to have all solutions to be rotated around their origin.

To shift the triangle, It has to be done through the connection the points ABC ' to the origin in such a way that the line is extended as we can seen in the diagram.

Read more on reflection here:

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Answer:

Step-by-step explanation:

P(4,3), Q(4,1), S(-1,3), R(-1,1)

[tex]Distance =\sqrt{(x_{2}-x_{1})^{2}+(y_{2} -y_{1})^{2}}\\\\PQ= \sqrt{(4-4)^{2}+(1-3)^{2}}\\\\=\sqrt{(-2)^{2}}\\\\\=\sqrt{4}\\\\=2 \ units\\\\\\QS=\sqrt{(-1-4)^{2}+(3-1)^{2}}\\\\=\sqrt{(-5)^{2}+(2)^{2}}\\\\=\sqrt{25+4}\\\\=\sqrt{29}\ units\\\\\\SR =\sqrt{(-1-[-1])^{2}+(1-3)^{2}}\\\\=\sqrt{(-1+1)^{2}+(-2)^{2}}\\\\=\sqrt{0+4}\\\\= \sqrt{4}\\\\= 2 \\\\\\PR = \sqrt{(-1-4)^{2}+(1-3)^{2}}\\\\=\sqrt{(-5)^{2}+(-2)^{2}}\\\\=\sqrt{25+4}\\\\=\sqrt{29}\\\\[/tex]

PQRS is a rectangle

Area= length *breadth

       = 2 * √29

       = 2√29 sq.units

Answer:

7 brown M&Ms.

Step-by-step explanation:

This question is not complete, but I will assume that the final question is how many brown M&Ms will be in this package.

0.14 × 52 is our equation.

The answer is 7.28. We cannot have .28 of a brown M&M in a package (unless you count the broken ones) so there will be, on average, 7 brown M&Ms in a package.

(again, the question is incomplete, so this may not be the answer)

Answer:

-2/3

Step-by-step explanation:

A rational number is a number that can be expressed by a fraction so when y add to fractions it’s a rational number.

Answer:

1) has bigger answer

Step-by-step explanation:

1)

solving parenthesis first we get

5 × 10

so, the answer = 50

2)

solving 3 × 5 first as we have to see multiplication first then addition

2 + 15 + 5

22

comparing both

50 > 22

so problem 1 has a bigger answer

Answer:

34m = c

Step-by-step explanation:

For every month (m) you pay 34 dollars. However many months youu use that service time 34 equals your total cost (c).

Answer:

[tex]let \: cost \: be \: { \bf{c}} \: and \: months \: be \: { \bf{n}} \\ { \bf{c \: \alpha \: n}} \\ { \bf{c = kn}} \\ 34 = (k \times 1) \\ k = 34 \: dollars \\ \\ { \boxed{ \bf{c = 34n}}}[/tex]

Answer: 5 ft i think so

Use the Pythagorean theorem

Answer:

a) Figure B and Figure C

b) Figure C

c) Figure A, Figure B, and Figure C

Step-by-step explanation:

a) Rectangles are shapes that have four sides, and four right angels. Right angles are angles that are 90 degrees.

The only shapes with four sides AND four right angles are Figure C and Figure B.

b) Squares are shapes with four EQUAL sides and four right angles. The only shape with four equal sides and four right angles is Figure C.

c) Parallelograms are any shapes with four sides. All of these shapes have four sides.

Hope this helps!

Answer:

B

Step-by-step explanation:

The coeffecients (I totally didn't spell that right) and variables match up.

Answer:

SECOND ONE

Step-by-step explanation: