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marissa [1.9K]
3 years ago
7

What is the y-intercept ofa line that has a slope of -3 and passes through (0,-7)​

Mathematics
1 answer:
Wewaii [24]3 years ago
5 0

Y intercept is -7

Look at the picture to see what I did

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There are sixteen ounces in a pound. 3/8 x 16/1= 48/8=6 oz
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Find the area of the region that lies inside the first curve and outside the second curve.
marishachu [46]

Answer:

Step-by-step explanation:

From the given information:

r = 10 cos( θ)

r = 5

We are to find the  the area of the region that lies inside the first curve and outside the second curve.

The first thing we need to do is to determine the intersection of the points in these two curves.

To do that :

let equate the two parameters together

So;

10 cos( θ) = 5

cos( θ) = \dfrac{1}{2}

\theta = -\dfrac{\pi}{3}, \ \  \dfrac{\pi}{3}

Now, the area of the  region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \  \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \  5^2 d \theta

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} 100 \ cos^2 \  \theta  d \theta - \dfrac{25}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \   d \theta

A = 50 \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  \dfrac{cos \ 2 \theta +1}{2}  \end {pmatrix} \ \ d \theta - \dfrac{25}{2}  \begin {bmatrix} \theta   \end {bmatrix}^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}

A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  {cos \ 2 \theta +1}  \end {pmatrix} \ \    d \theta - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}}    \ \ - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{2 \pi}{3} \end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin (\dfrac{2 \pi}{3} )}{2}+\dfrac{\pi}{3} - \dfrac{ sin (\dfrac{-2\pi}{3}) }{2}-(-\dfrac{\pi}{3})  \end {bmatrix} - \dfrac{25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} +   \dfrac{\pi}{3}  \end {bmatrix}- \dfrac{ 25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3}   \end {bmatrix}- \dfrac{ 25 \pi}{3}

A =    \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}

The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.

Download docx
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Find the number of real number solutions for the equation. x2 + 5x + 7 = 0
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What is the simplified form of 24 y to the eighth power over 15 x to the fifth power divided by 8 y to the fourth power over 4 x
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the answer to the question

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You purchase 5 pounds of apples and 2 pounds of oranges for $9.Your friend purchases 5 pounds of apples and 6 pounds of oranges
babymother [125]

Answer:

x = 1 and y = 2

Step-by-step explanation:

Let apples are represented by x

and let oranges are represented by y

You purchase 5 pounds of apples and 2 pounds of oranges for $9. This line in equation format can be written as:

5x + 2y = 9

Your friend purchases 5 pounds of apples and 6 pounds of oranges for $17.

This line in equation format can be written as:

5x + 6y = 17

Now we have two equations:

5x + 2y = 9 -> eq (i)

5x + 6y = 17 -> eq(ii)

We can solve these equations to find the value of x and y.

Subtracting eq(i) from eq(ii)

5x + 6y = 17

5x + 2y = 9

-     -         -

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0+4y= 8

=> 4y = 8

y= 8/4

y = 2

Now, putting value of y in eq (i)

5x + 2y = 9

5x +2(2) = 9

5x +4 = 9

5x = 9-4

5x = 5

x = 1

so, x = 1 and y = 2

8 0
3 years ago
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