S = sally C = Chris J = Joe
sally is one year older than 3 times the age of joe, so
s = 3j + 1
chris is 5 years younger than 6 times the age of jo, so
c = 6j - 5
sally and chris are the same age so
s = c so
3j + 1 = 6j - 5
-3j -3j
1 = 3j - 5
+5 +5
6 = 3j or
3j = 6
3j/3 = 6/3
j = 2
check
s = c
3(2) +1 = 6(2) - 5
6 + 1 = 12 - 5
7 = 7 correct✅
Answer:
54π
Step-by-step explanation:
To find the volume of a cylinder, you multiply the area of the base times the height.
The formula for the area of a circle is πr², so the base is 9π units².
Multiplying that by the height (6 units) gives you 54π units³.
Theres no graphs but this is how the function looks like
![\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} \stackrel{\textit{we'll use this one}}{y=a(x- h)^2+ k}\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{2}{ h},\stackrel{-1}{ k}) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=2\\ k=-1 \end{cases}\implies y=a(x-2)^2-1 \\\\\\ \textit{we also know that } \begin{cases} y=0\\ x=5 \end{cases}\implies 0=a(5-2)^2-1\implies 1=9a \\\\\\ \cfrac{1}{9}=a\qquad therefore\qquad \boxed{y=\cfrac{1}{9}(x-2)^2-1}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%5Ctextit%7Bparabola%20vertex%20form%7D%20%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20%5Cstackrel%7B%5Ctextit%7Bwe%27ll%20use%20this%20one%7D%7D%7By%3Da%28x-%20h%29%5E2%2B%20k%7D%5C%5C%5C%5C%20x%3Da%28y-%20k%29%5E2%2B%20h%20%5Cend%7Barray%7D%20%5Cqquad%5Cqquad%20vertex~~%28%5Cstackrel%7B2%7D%7B%20h%7D%2C%5Cstackrel%7B-1%7D%7B%20k%7D%29%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cbegin%7Bcases%7D%20h%3D2%5C%5C%20k%3D-1%20%5Cend%7Bcases%7D%5Cimplies%20y%3Da%28x-2%29%5E2-1%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Bwe%20also%20know%20that%20%7D%20%5Cbegin%7Bcases%7D%20y%3D0%5C%5C%20x%3D5%20%5Cend%7Bcases%7D%5Cimplies%200%3Da%285-2%29%5E2-1%5Cimplies%201%3D9a%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B1%7D%7B9%7D%3Da%5Cqquad%20therefore%5Cqquad%20%5Cboxed%7By%3D%5Ccfrac%7B1%7D%7B9%7D%28x-2%29%5E2-1%7D)
now, let's expand the squared term to get the standard form of the quadratic.
![\bf y=\cfrac{1}{9}(x-2)^2-1\implies y=\cfrac{1}{9}(x^2-4x+4)-1 \\\\\\ y=\cfrac{1}{9}x^2-\cfrac{4}{9}x+\cfrac{4}{9}-1\implies \stackrel{its~coefficient}{y=\stackrel{\downarrow }{\cfrac{1}{9}}x^2-\cfrac{4}{9}x-\cfrac{5}{9}}](https://tex.z-dn.net/?f=%5Cbf%20y%3D%5Ccfrac%7B1%7D%7B9%7D%28x-2%29%5E2-1%5Cimplies%20y%3D%5Ccfrac%7B1%7D%7B9%7D%28x%5E2-4x%2B4%29-1%20%5C%5C%5C%5C%5C%5C%20y%3D%5Ccfrac%7B1%7D%7B9%7Dx%5E2-%5Ccfrac%7B4%7D%7B9%7Dx%2B%5Ccfrac%7B4%7D%7B9%7D-1%5Cimplies%20%5Cstackrel%7Bits~coefficient%7D%7By%3D%5Cstackrel%7B%5Cdownarrow%20%7D%7B%5Ccfrac%7B1%7D%7B9%7D%7Dx%5E2-%5Ccfrac%7B4%7D%7B9%7Dx-%5Ccfrac%7B5%7D%7B9%7D%7D)