A. Alright, we want to multiply one equation by a constant to make it cancel out with the second. Since the first equation has a "blank" y, let's multiply the first equation by <em>2</em>.

3x-y=0 → 2(3x-y=0) = 6x - 2y = 0

5x+2y=22

The answer for this part would be: 6x - 2y = 0 and 5x + 2y = 22

B. So now we combine them:

6x - 2y = 0

+ + +

5x + 2y = 22

= = =

11x + 0 = 22 ← The answer

C. Now that we have the equation 11x = 22, we solve for x

11x = 22 ← Divide both sides by 11

x = 2 ← The answer

D. Now that we have x=2, we plug that back in to 5x+2y=22 and solve for y:

5(2)+2y = 22

10 + 2y = 22

2y = 12

y = 6

<u>Therefore, the solution to this problem is x = 2 and y = 6</u>

5 0

3 years ago

Find the area of the region which is inside the polar curve r=5sin(θ) but outside r=4. Round your answer to four decimal places

natka813 [3]

The area of the region which is inside the polar curve r = 5 sinθ but outside r = 4 will be 3.75 square units.

<h3>What is an area bounded by the curve?</h3>

When the two curves intersect then they bound the region is known as the area bounded by the curve.

The area of the region which is inside the polar curve r = 5 sinθ but outside r = 4 will be

Then the intersection point will be given as

$\rm 5 \sin \theta = 4\\\\\theta = 0.927 , 2.214$

Then by the integration, we have

$\rightarrow \dfrac{1}{2} \times \int _{0.927}^{2.214}[ (5 \sin \theta)^2 - 4^2] d\theta \\\\\\\rightarrow \dfrac{1}{2} \times \int _{0.927}^{2.214} [25\sin ^2 \theta - 16] d\theta \\\\\\\rightarrow \dfrac{1}{2} \times \int _{0.927}^{2.214} [ \dfrac{25}{2}(1 - \cos 2\theta ) - 16] d\theta \\$

$\rightarrow \dfrac{1}{2} [\dfrac{25 \theta }{5} - \dfrac{25 \cos 2\theta }{2} - 16\theta]_{0.927}^{2.214} \\\\\\\rightarrow \dfrac{1}{2} [\dfrac{25(2.214 - 0.927) }{5} - \dfrac{25 (\cos 2\times 2.214 - \cos 2\times 0.927) }{2} - 16(2.214 - 0.927]\\$