let's first off take a peek at those values.
let's say the point with those coordinates is point C, so C is 3/10 of the way from A to B.
meaning, we take the segment AB and cut it in 10 equal pieces, AC takes 3 pieces, and CB takes 7 pieces, namely AC and CB are at a 3:7 ratio.
![\bf ~~~~~~~~~~~~\textit{internal division of a line segment} \\\\\\ A(-4,-8)\qquad B(11,7)\qquad \qquad \stackrel{\textit{ratio from A to B}}{3:7} \\\\\\ \cfrac{A\underline{C}}{\underline{C} B} = \cfrac{3}{7}\implies \cfrac{A}{B} = \cfrac{3}{7}\implies 7A=3B\implies 7(-4,-8)=3(11,7)\\\\[-0.35em] ~\dotfill\\\\ C=\left(\frac{\textit{sum of "x" values}}{\textit{sum of ratios}}\quad ,\quad \frac{\textit{sum of "y" values}}{\textit{sum of ratios}}\right)\\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Binternal%20division%20of%20a%20line%20segment%7D%0A%5C%5C%5C%5C%5C%5C%0AA%28-4%2C-8%29%5Cqquad%20B%2811%2C7%29%5Cqquad%0A%5Cqquad%20%5Cstackrel%7B%5Ctextit%7Bratio%20from%20A%20to%20B%7D%7D%7B3%3A7%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7BA%5Cunderline%7BC%7D%7D%7B%5Cunderline%7BC%7D%20B%7D%20%3D%20%5Ccfrac%7B3%7D%7B7%7D%5Cimplies%20%5Ccfrac%7BA%7D%7BB%7D%20%3D%20%5Ccfrac%7B3%7D%7B7%7D%5Cimplies%207A%3D3B%5Cimplies%207%28-4%2C-8%29%3D3%2811%2C7%29%5C%5C%5C%5C%5B-0.35em%5D%0A~%5Cdotfill%5C%5C%5C%5C%0AC%3D%5Cleft%28%5Cfrac%7B%5Ctextit%7Bsum%20of%20%22x%22%20values%7D%7D%7B%5Ctextit%7Bsum%20of%20ratios%7D%7D%5Cquad%20%2C%5Cquad%20%5Cfrac%7B%5Ctextit%7Bsum%20of%20%22y%22%20values%7D%7D%7B%5Ctextit%7Bsum%20of%20ratios%7D%7D%5Cright%29%5C%5C%5C%5C%5B-0.35em%5D%0A~%5Cdotfill)
![\bf C=\left(\cfrac{(7\cdot -4)+(3\cdot 11)}{3+7}\quad ,\quad \cfrac{(7\cdot -8)+(3\cdot 7)}{3+7}\right) \\\\\\ C=\left( \cfrac{-28+33}{10}~~,~~\cfrac{-56+21}{10} \right)\implies C=\left( \cfrac{5}{10}~~,~~\cfrac{-35}{10} \right) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill C=\left( \frac{1}{2}~,~-\frac{7}{2} \right)~\hfill](https://tex.z-dn.net/?f=%5Cbf%20C%3D%5Cleft%28%5Ccfrac%7B%287%5Ccdot%20-4%29%2B%283%5Ccdot%2011%29%7D%7B3%2B7%7D%5Cquad%20%2C%5Cquad%20%5Ccfrac%7B%287%5Ccdot%20-8%29%2B%283%5Ccdot%207%29%7D%7B3%2B7%7D%5Cright%29%0A%5C%5C%5C%5C%5C%5C%0AC%3D%5Cleft%28%20%5Ccfrac%7B-28%2B33%7D%7B10%7D~~%2C~~%5Ccfrac%7B-56%2B21%7D%7B10%7D%20%5Cright%29%5Cimplies%20C%3D%5Cleft%28%20%5Ccfrac%7B5%7D%7B10%7D~~%2C~~%5Ccfrac%7B-35%7D%7B10%7D%20%5Cright%29%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A~%5Chfill%20C%3D%5Cleft%28%20%5Cfrac%7B1%7D%7B2%7D~%2C~-%5Cfrac%7B7%7D%7B2%7D%20%5Cright%29~%5Chfill)
wdym: 14 of the total cake
Answer:
D
Step-by-step explanation:
Answer:
Part a) The lateral area is 
Part b) The area of the two bases together is 
Part c) The surface area is 
Step-by-step explanation:
we know that
The surface area of a right cylinder is equal to

where
LA is the lateral area
B is the area of the base of cylinder
we have


Part a) Find the lateral area
The lateral area is equal to

substitute the values


Part b) Find the area of the two bases together
The area of the base B is equal to

so
the area of the two bases together is

Part c) Find the surface area of the cylinder

we have


substitute

Answer:420
Step-by-step explanation: