412 pounds for him + 1 pound for his friend = 413 total pounds x $0.80 per pound = $330.40
Sorry, I don’t see this as an answer choice.
Notice the picture below, with an squared area, running along the ellipse, perpendicular to the x-axis
so
![\bf x^2+25y^2=25\implies y^2=\cfrac{25-x^2}{25}\implies y=\sqrt{1-\cfrac{x^2}{25}} \\\\\\ \textit{that will be half of one side, notice in the picture}\\ \textit{so, a full side will be }2\sqrt{1-\cfrac{x^2}{25}} \\\\\\ \textit{now, the area of a square is }length\times width \\\\\\ but\quad \begin{cases} length=2\sqrt{1-\cfrac{x^2}{25}}\\\\ width=2\sqrt{1-\cfrac{x^2}{25}} \end{cases} \\\\\\ \textit{thus the area of it is }\left( 2\sqrt{1-\cfrac{x^2}{25}} \right)\left( 2\sqrt{1-\cfrac{x^2}{25}} \right)](https://tex.z-dn.net/?f=%5Cbf%20x%5E2%2B25y%5E2%3D25%5Cimplies%20y%5E2%3D%5Ccfrac%7B25-x%5E2%7D%7B25%7D%5Cimplies%20y%3D%5Csqrt%7B1-%5Ccfrac%7Bx%5E2%7D%7B25%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Bthat%20will%20be%20half%20of%20one%20side%2C%20notice%20in%20the%20picture%7D%5C%5C%0A%5Ctextit%7Bso%2C%20a%20full%20side%20will%20be%20%7D2%5Csqrt%7B1-%5Ccfrac%7Bx%5E2%7D%7B25%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Bnow%2C%20the%20area%20of%20a%20square%20is%20%7Dlength%5Ctimes%20width%0A%5C%5C%5C%5C%5C%5C%0Abut%5Cquad%20%0A%5Cbegin%7Bcases%7D%0Alength%3D2%5Csqrt%7B1-%5Ccfrac%7Bx%5E2%7D%7B25%7D%7D%5C%5C%5C%5C%0Awidth%3D2%5Csqrt%7B1-%5Ccfrac%7Bx%5E2%7D%7B25%7D%7D%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Bthus%20the%20area%20of%20it%20is%20%7D%5Cleft%28%202%5Csqrt%7B1-%5Ccfrac%7Bx%5E2%7D%7B25%7D%7D%20%5Cright%29%5Cleft%28%202%5Csqrt%7B1-%5Ccfrac%7Bx%5E2%7D%7B25%7D%7D%20%5Cright%29)
![\bf or\quad \left( 2\sqrt{1-\cfrac{x^2}{25}} \right)^2](https://tex.z-dn.net/?f=%5Cbf%20or%5Cquad%20%20%20%5Cleft%28%202%5Csqrt%7B1-%5Ccfrac%7Bx%5E2%7D%7B25%7D%7D%20%5Cright%29%5E2)
now... how far is it from the left-side to the right-side of the ellipse?
well, that's the major axis of it
let's see which one is that in this case
![\bf x^2+25y^2=25\implies\cfrac{x^2}{25}+\cfrac{25y^2}{25}=1 \\\\\\ \cfrac{(x-0)^2}{5^2}+\cfrac{(y-0)^2}{1^2}=1\\\\ -----------------------------\\\\ \cfrac{(x-{{ h}})^2}{{{ a}}^2}+\cfrac{(y-{{ k}})^2}{{{ b}}^2}=1 \ \ center\ ({{ h}},{{ k}})\qquad vertices\ ({{ h}}\pm a, {{ k}})\\\\ -----------------------------\\\\ \begin{cases} a=5\\ b=1\\ h=0\\ k=0 \end{cases}\implies vertices\to (0\pm 5,0)\to (\pm 5,0)](https://tex.z-dn.net/?f=%5Cbf%20x%5E2%2B25y%5E2%3D25%5Cimplies%5Ccfrac%7Bx%5E2%7D%7B25%7D%2B%5Ccfrac%7B25y%5E2%7D%7B25%7D%3D1%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B%28x-0%29%5E2%7D%7B5%5E2%7D%2B%5Ccfrac%7B%28y-0%29%5E2%7D%7B1%5E2%7D%3D1%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0A%5Ccfrac%7B%28x-%7B%7B%20h%7D%7D%29%5E2%7D%7B%7B%7B%20a%7D%7D%5E2%7D%2B%5Ccfrac%7B%28y-%7B%7B%20k%7D%7D%29%5E2%7D%7B%7B%7B%20b%7D%7D%5E2%7D%3D1%0A%0A%5C%20%5C%20center%5C%20%28%7B%7B%20h%7D%7D%2C%7B%7B%20k%7D%7D%29%5Cqquad%0A%20vertices%5C%20%28%7B%7B%20h%7D%7D%5Cpm%20a%2C%20%7B%7B%20k%7D%7D%29%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0A%0A%5Cbegin%7Bcases%7D%0Aa%3D5%5C%5C%0Ab%3D1%5C%5C%0Ah%3D0%5C%5C%0Ak%3D0%0A%5Cend%7Bcases%7D%5Cimplies%20vertices%5Cto%20%280%5Cpm%205%2C0%29%5Cto%20%28%5Cpm%205%2C0%29)
so the ellipse runs from -5 to 5 over the x-axis
thus the area will be
Answer:
option 2 with radius of 1.4 in, and height of 1.2 in.
Step-by-step explanation:
If two cylinders are similar, the ratio of one cylinder's radius to its height must be the same as that of the other.
To know which cylinder is similar to the given cylinder with radius 2.8 in and height of 2.4 in, find the ratio, and compare with the ratio of the options provided. The option with the same ratio, is the cylinder that is similar.
This,
The given cylinder => radius : height = ![\frac{2.8}{2.4} = \frac{0.7}{0.6} = \frac{7}{6}](https://tex.z-dn.net/?f=%20%5Cfrac%7B2.8%7D%7B2.4%7D%20%3D%20%5Cfrac%7B0.7%7D%7B0.6%7D%20%3D%20%5Cfrac%7B7%7D%7B6%7D%20)
First option:
Radius : height = ![\frac{1.8}{1.4} = \frac{0.9}{0.7} = \frac{9}{7}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1.8%7D%7B1.4%7D%20%3D%20%5Cfrac%7B0.9%7D%7B0.7%7D%20%3D%20%5Cfrac%7B9%7D%7B7%7D%20)
Second option:
Radius : height = ![\frac{1.4}{1.2} = \frac{0.7}{0.6} = \frac{7}{6}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1.4%7D%7B1.2%7D%20%3D%20%5Cfrac%7B0.7%7D%7B0.6%7D%20%3D%20%5Cfrac%7B7%7D%7B6%7D%20)
Third option:
Radius : height = ![\frac{5.6}{4.2} = \frac{0.8}{0.6} = \frac{0.4}{0.3} = \frac{4}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B5.6%7D%7B4.2%7D%20%3D%20%5Cfrac%7B0.8%7D%7B0.6%7D%20%3D%20%5Cfrac%7B0.4%7D%7B0.3%7D%20%3D%20%5Cfrac%7B4%7D%7B3%7D)
Fourth option:
Radius : height = ![\frac{2.4}{2.8} = \frac{0.6}{0.7} = \frac{6}{7}](https://tex.z-dn.net/?f=%20%5Cfrac%7B2.4%7D%7B2.8%7D%20%3D%20%5Cfrac%7B0.6%7D%7B0.7%7D%20%3D%20%5Cfrac%7B6%7D%7B7%7D%20)
The correct option with the cylinder that is similar with the given cylinder is option 2 with radius of 1.4 in, and height of 1.2 in.
Answer:
3
Step-by-step explanation:
3² + 4² = 5²
Hope this helps :)
F(t)= $100 x h + 300
A reasonable domain is (1,2,3) and the range is ($400,$500,$600)