The gross profit margin is calculated using the following rule:
gross profit margin = total profit / total sales
Now, we need to get the values of total profit and total sale:
total profit = <span>9*8-(40+8)=24$
total sales = 9*8 = 72$
Now, we will substitute in the above equation:
gross profit margin = 24/72 = 1/3 = 0.3333334
% = 0.33333334*100 = 33.3334%</span>
A) Add three <em>line</em> segments (AD, CF, BE) to the <em>regular</em> hexagon.
B) The area of each triangle of the <em>regular</em> hexagon is 35.1 in².
C) The area of the <em>regular</em> hexagon is 210.6 in².
<h3>How to calculate the area of a regular hexagon</h3>
In geometry, regular hexagons are formed by six <em>regular</em> triangles with a common vertex. We decompose the hexagon in six <em>equilateral</em> triangles by adding three <em>line</em> segments (AD, CF, BE).The area of each triangle is found by the following equation:
A = 0.5 · (9 in) · (7.8 in)
A = 35.1 in²
And the area of the <em>regular</em> polygon is six times the former result, that is, 210.6 square inches.
To learn more on polygons: brainly.com/question/17756657
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The initial membership fee before any months are considered is 35.
Substitute 0 with x because no months have been considered. Like what was previously said, y is the total cost, so:
Y=12(0)+35
12 x 0 = 0
Y=0+35
Y=35
Answer:
n to the power of -2
Step-by-step explanation:
i counted all the others and this is the only one that doesn't add up to 16. the last answer is -16 not 16 so there you go.
Another triple integral. We're integrating over the interior of the sphere
![x^2+y^2+z^2=2^2](https://tex.z-dn.net/?f=x%5E2%2By%5E2%2Bz%5E2%3D2%5E2)
Let's do the outer integral over z. z stays within the sphere so it goes from -2 to 2.
For the middle integral we have
![y^2=4-x^2-z^2](https://tex.z-dn.net/?f=y%5E2%3D4-x%5E2-z%5E2)
x is the inner integral so at this point we conservatively say its zero. That means y goes from
and ![+\sqrt{4-z^2}](https://tex.z-dn.net/?f=%2B%5Csqrt%7B4-z%5E2%7D)
Similarly the inner integral x goes between ![\pm-\sqrt{4-y^2-z^2}](https://tex.z-dn.net/?f=%5Cpm-%5Csqrt%7B4-y%5E2-z%5E2%7D)
So we rewrite the integral
![\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%7B-2%7D%5E%7B2%7D%20%5Cint_%7B-%5Csqrt%7B4-z%5E2%7D%7D%5E%7B%5Csqrt%7B4-z%5E2%7D%7D%20%5Cint_%7B-%5Csqrt%7B4-y%5E2-z%5E2%7D%7D%5E%7B%5Csqrt%7B4-y%5E2-z%5E2%7D%7D%20%28x%5E2%2Bxy%2By%5E2%29dx%20%5C%3B%20dy%20%5C%3B%20dz)
Let's work on the inner one,
![\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_%7B-%5Csqrt%7B4-y%5E2-z%5E2%7D%7D%5E%7B%5Csqrt%7B4-y%5E2-z%5E2%7D%7D%20%28x%5E2%2Bxy%2By%5E2%29dz)
There's no z in the integrand, so we treat it as a constant.
![=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}](https://tex.z-dn.net/?f=%3D%28x%5E2%2Bxy%2By%5E2%29z%20%5Cbigg%7C_%7Bz%3D-%5Csqrt%7B4-y%5E2-z%5E2%7D%7D%5E%7Bz%3D%5Csqrt%7B4-y%5E2-z%5E2%7D%7D)
So the middle integral is
I gotta go so I'll stop here, sorry.