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3 years ago

Determine the intercepts of the line.

Mathematics

1 answer:

jeyben [28]3 years ago

5 0

Answer:

y- intercept = - 13 , x- intercept = $\frac{13}{5}$

Step-by-step explanation:

To find the y- intercept, substitute x = 0 into the equation and solve for y

y = 5(0) - 13 = 0 - 13 = - 13 ← y- intercept

To find the x- intercept substitute y = 0 into the equation and solve for x

0 = 5x - 13 ( add 13 to both sides )

13 = 5x ( divide both sides by 5 )

$\frac{13}{5}$ = x

x- intercept = $\frac{13}{5}$ = 2 $\frac{3}{5}$ = 2.6

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professor190 [17]

Hello! i can do the first word problem

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3 years ago

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

7 0

3 years ago

In a dance competition, a participant has to score a total of at least 30 points in the first four rounds combined to move on to

brilliants [131]

If he got 5 points in the first round and he needed at least thirty he would score 9 points in the next three rounds

8 0

3 years ago

what fraction is equivalent to -3/2? i’m going to keep writing because it has to be 20 words. Anyways thank you if you answer th

Alinara [238K]

A fraction that is equivalent to -3/2 would be -9/6

A great form to prove this is by using a calculator and dividing -9 by 6 this should get you the answer of 1.5!

Hope this helps!

7 0

3 years ago

Lol plsss help me :((

patriot [66]

Answer:

The choice two;

$- \sqrt{3}$

Step-by-step explanation:

$\sqrt{27} - 4 \sqrt{3} \\ 3 \sqrt{3} - 4 \sqrt{3} = - 1 \sqrt{3} = - \sqrt{3}$

6 0

3 years ago