First you will divide 100 by 50 you get 2. then you subrtact 38 from 50. you get 12. then since 100 is twice the amount as 50, you will multiply 12 by 2. you get 24. that is your answer
Let x be the unknown number
x×55%=121
x=121÷55%
x=220
So 121 is 55% of 220
Answer:
The change in volume is estimated to be 17.20 ![\rm{in^3}](https://tex.z-dn.net/?f=%5Crm%7Bin%5E3%7D)
Step-by-step explanation:
The linearization or linear approximation of a function
is given by:
where
is the total differential of the function evaluated in the given point.
For the given function, the linearization is:
![V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh](https://tex.z-dn.net/?f=V%28R_0%2BdR%2C%20h_0%2Bdh%29%20%3D%20V%28R_0%2C%20h_0%29%20%2B%20%5Cfrac%7B%5Cpartial%20V%28R_0%2C%20h_0%29%7D%7B%5Cpartial%20R%7DdR%20%2B%20%5Cfrac%7B%5Cpartial%20V%28R_0%2C%20h_0%29%7D%7B%5Cpartial%20h%7Ddh)
Taking
inches and
inches and evaluating the partial derivatives we obtain:
![V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh\\V(R, h) = V(R_0, h_0) + (\frac{2 h \pi r}{3} + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh](https://tex.z-dn.net/?f=V%28R_0%2BdR%2C%20h_0%2Bdh%29%20%3D%20V%28R_0%2C%20h_0%29%20%2B%20%5Cfrac%7B%5Cpartial%20V%28R_0%2C%20h_0%29%7D%7B%5Cpartial%20R%7DdR%20%2B%20%5Cfrac%7B%5Cpartial%20V%28R_0%2C%20h_0%29%7D%7B%5Cpartial%20h%7Ddh%5C%5CV%28R%2C%20h%29%20%3D%20V%28R_0%2C%20h_0%29%20%2B%20%28%5Cfrac%7B2%20h%20%5Cpi%20r%7D%7B3%7D%20%20%2B%202%20%5Cpi%20r%5E2%29dR%20%2B%20%28%5Cfrac%7B%5Cpi%20r%5E2%7D%7B3%7D%20%29dh)
substituting the values and taking
and
inches we have:
![V(R_0+dR, h_0+dh) =V(R_0, h_0) + (\frac{2 h \pi r}{3} + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh\\V(1.5+0.1, 3+0.3) =V(1.5, 3) + (\frac{2 \cdot 3 \pi \cdot 1.5}{3} + 2 \pi 1.5^2)\cdot 0.1 + (\frac{\pi 1.5^2}{3} )\cdot 0.3\\V(1.5+0.1, 3+0.3) = 17.2002\\\boxed{V(1.5+0.1, 3+0.3) \approx 17.20}](https://tex.z-dn.net/?f=V%28R_0%2BdR%2C%20h_0%2Bdh%29%20%3DV%28R_0%2C%20h_0%29%20%2B%20%28%5Cfrac%7B2%20h%20%5Cpi%20r%7D%7B3%7D%20%20%2B%202%20%5Cpi%20r%5E2%29dR%20%2B%20%28%5Cfrac%7B%5Cpi%20r%5E2%7D%7B3%7D%20%29dh%5C%5CV%281.5%2B0.1%2C%203%2B0.3%29%20%3DV%281.5%2C%203%29%20%2B%20%28%5Cfrac%7B2%20%5Ccdot%203%20%5Cpi%20%5Ccdot%201.5%7D%7B3%7D%20%20%2B%202%20%5Cpi%201.5%5E2%29%5Ccdot%200.1%20%2B%20%28%5Cfrac%7B%5Cpi%201.5%5E2%7D%7B3%7D%20%29%5Ccdot%200.3%5C%5CV%281.5%2B0.1%2C%203%2B0.3%29%20%3D%2017.2002%5C%5C%5Cboxed%7BV%281.5%2B0.1%2C%203%2B0.3%29%20%5Capprox%2017.20%7D)
Therefore the change in volume is estimated to be 17.20 ![\rm{in^3}](https://tex.z-dn.net/?f=%5Crm%7Bin%5E3%7D)
The simplified equation is 2ab, so the coefficient would be 2.
Fourth test mark should be 86
85 + 82 + 75 + 86 = 328
328 divide by 4 (as there are four tests) = 82