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2 years ago

Can someone please help me?

Mathematics

1 answer:

Allisa [31]2 years ago

7 0

Answer: you got this

Step-by-step explanation: YOU WILL NEVER SUCCEED IF YOU DON’T TRY

search it up if you really need help

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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

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

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

On solving, we have

$\rightarrow \dfrac{1}{2} \times 7.499\\\\\rightarrow 3.75$

Thus, the area of the region is 3.75 square units.

More about the area bounded by the curve link is given below.

brainly.com/question/24563834

#SPJ4

5 0

2 years ago

Pearl lives in Atlanta but spent 1⁄6 of the year in New York and 4⁄9 of the year in Los Angeles. What fraction of the year did s

solong [7]

She spent 5/6th of nights away from atlanta

6 0

3 years ago

I need help please 25 points for whoever helps me

34kurt

Yuto is correct because he isolated the variable correctly and reversed the inequality symbol.

x < 3

3 0

3 years ago

A gardener already has four and one over 2 ft of fencing in his garage. He wants to fence in a square garden for his flowers. Th

NikAS [45]

<h2>Answer:</h2>

$6\frac{1}{2}$ <u>, "six and one over two ft".</u>

<h2>Step-by-step explanation:</h2>

Let's considerate the fact that the garden has a <u>square shape</u>.

<h3>1. Finding values of interest.</h3>

Amount of fence that the gardener already has: $4\frac{1}{2}$ ft.

Length of one side: $2\frac{3}{4}$ ft.

If one side measures $2\frac{3}{4}$ ft, and the square garden has 4 sides of equal length, because it's a square, then we must multiply the measure of one side by 4 to find the total length of fence needed:

$4*(2\frac{3}{4})=\\ \\4*(2+\frac{3}{4})=\\ \\(4*2)+(4*\frac{3}{4})=\\ \\8+(\frac{12}{4} )=\\ \\8+3=\\ \\11$

The gardener already has $4\frac{1}{2}$ , which equals $4 + \frac{1}{2}$ . Hence, the difference between the amount needed and the amount that the gardeneralready has will give us the remaining amount required. Let's do that: