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3 years ago

A regular octagon has an apothem measuring 10 in. and a perimeter of 66.3 in.

Mathematics

2 answers:

Stells [14]3 years ago

8 0

Answer:

332 in.2

Step-by-step explanation:

ON MY LEFT CHEECK THIS RIGHTTTTT!!

POOH SHIESTY DATS MY DAWG BUT POOH U KNOW IM RLLY SHIESTEHHHHHH

Alexxandr [17]3 years ago

3 0

Answer:

332 in.2

Step-by-step explanation:

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Please see attached below. <br> Please explain.

vivado [14]

Answer:

15 rentals

Step-by-step explanation:

You can (and may be expected to) set up an equation that equates the total cost at one store to the total cost at the other store. When you work through the solution of this equation, you find that the "break even" number of rentals is the ratio of the difference in fixed cost (setup fee) to the difference in per-use cost (rental charge).

Here, that ratio is ...

(15.00 -7.50)/(2.25 -1.75) = 7.50/0.50 = 15

15 rentals will make the total costs the same.

5 0

3 years ago

Read 2 more answers

Two and one third minus two thirds

Savatey [412]

The answer to 2 1/3 - 2/3 is 1 2/3

4 0

2 years ago

What’s 3/4 of 2/5 ?<br><br> ** I WILL MAKE UP BRAIN LIST OR SOMETHING

Yanka [14]

Answer:

$\frac{3}{10}$

Step-by-step explanation:

$= \frac{3}{4} \times \frac{2}{5}$

$= \frac{3 \times 2}{4 \times 5}$

$= \frac{6}{20}$

$= \frac{6 \div 2}{20 \div 2}$

$= \frac{3}{10}$

7 0

3 years ago

Three times the sum of 4 and a number is twice the<br> number. Find the number.

zavuch27 [327]

Step-by-step explanation:

let number be x

3(4+x)=2x

12 + 3x = 2x

12 = 2x - 3x

12 = -x

x= -12

3 0

3 years ago

Find the area of the part of the plane 3x 2y z = 6 that lies in the first octant.

gavmur [86]

The area of the part of the plane 3x 2y z = 6 that lies in the first octant is mathematically given as

A=3 √(4) units ^2

<h3>What is the area of the part of the plane 3x 2y z = 6 that lies in the first octant.?</h3>

Generally, the equation for is mathematically given as

The Figure is the x-y plane triangle formed by the shading. The formula for the surface area of a z=f(x, y) surface is as follows:

$A=\iint_{R_{x y}} \sqrt{f_{x}^{2}+f_{y}^{2}+1} d x d y(1)$

The partial derivatives of a function are f x and f y.

$\begin{aligned}&Z=f(x)=6-3 x-2 y \\&=\frac{\partial f(x)}{\partial x}=-3 \\&=\frac{\partial f(y)}{\partial y}=-2\end{aligned}$

When these numbers are plugged into equation (1) and the integrals are given bounds, we get:

$&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{(-3)^{2}+(-2)^2+1dxdy} \\\\&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{14} d x d y \\\\&=\sqrt{14} \int_{0}^{2}[y]_{0}^{3-\frac{3}{2} x} d x d y \\\\&=\sqrt{14} \int_{0}^{2}\left[3-\frac{3}{2} x\right] d x \\\\$

$&=\sqrt{14}\left[3 x-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3.2-\frac{3}{2} \cdot \frac{1}{2} \cdot 3^{2}\right] \\\\&=3 \sqrt{14} \text { units }{ }^{2}$

In conclusion, the area is

A=3 √4 units ^2