CAN SOMEONE PLEASE HELP ME !! ONLY NUMBER 27

Mathematics

1 answer:

Yuki888 [10]2 years ago

5 0

Answer:

17.75

Step-by-step explanation:

1- You need multiply 1.75 by 9 because the first day is an uneven amount.

2- Then we will take your answer which is 15.75 and the 2 miles she whatever ran something like that, and add it.

3- Your final answer is 17.75.

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Surface area and volume

Wittaler [7]

Triangular prism
length of slanted side:
sqrt(3^2 + 4^2) = 5
SA = 2(1/2)(8)(3) + 8(12) + 2(5)(12) =240
V = (1/2)(8)(3)(12) = 144

cuboid
SA = 2(7x6) + 2(7x5) + 2(6x5) = 214
V = 7x6x5 = 210

6 0

2 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

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

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