Answer:
Yes
Step-by-step explanation:
They are just in the opposite side
You would use Pythagorean theorem to help u out if u have 2 sides u should be able to do it.
<h3>
Answer: 29</h3>
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Work Shown:
Solve the second equation for f(b)
2f(b) = 28
f(b) = 28/2
f(b) = 14
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Plug in x = b and solve for b
f(x) = 3x-1
f(b) = 3b-1 ... replace every x with b
14 = 3b-1 ..... plug in f(b) = 14
3b-1 = 14
3b = 14+1
3b = 15
b = 15/3
b = 5
-----------
f(x) = 3x-1
f(2b) = 3(2b)-1 .... replace every x with 2b
f(2b) = 3(2*5)-1 .... plug in b = 5
f(2b) = 3*10-1
f(2b) = 30-1
f(2b) = 29
Check the picture below, so pretty much reaches its maximum height at the vertex, now let's take a peek at the equation above hmmmm
![~~~~~~\textit{vertical parabola vertex form} \\\\ y=a(x- h)^2+ k\qquad \begin{cases} \stackrel{vertex}{(h,k)}\\\\ \stackrel{"a"~is~negative}{op ens~\cap}\qquad \stackrel{"a"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] ~\dotfill\\\\ h(t)=-16(t ~~ - ~~ \stackrel{h}{5})^2~~ + ~~\stackrel{k}{116}~\hfill \underset{maximum~height}{\stackrel{vertex}{(5~~,~~\underset{\uparrow }{116})}}](https://tex.z-dn.net/?f=~~~~~~%5Ctextit%7Bvertical%20parabola%20vertex%20form%7D%20%5C%5C%5C%5C%20y%3Da%28x-%20h%29%5E2%2B%20k%5Cqquad%20%5Cbegin%7Bcases%7D%20%5Cstackrel%7Bvertex%7D%7B%28h%2Ck%29%7D%5C%5C%5C%5C%20%5Cstackrel%7B%22a%22~is~negative%7D%7Bop%20ens~%5Ccap%7D%5Cqquad%20%5Cstackrel%7B%22a%22~is~positive%7D%7Bop%20ens~%5Ccup%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20h%28t%29%3D-16%28t%20~~%20-%20~~%20%5Cstackrel%7Bh%7D%7B5%7D%29%5E2~~%20%2B%20~~%5Cstackrel%7Bk%7D%7B116%7D~%5Chfill%20%5Cunderset%7Bmaximum~height%7D%7B%5Cstackrel%7Bvertex%7D%7B%285~~%2C~~%5Cunderset%7B%5Cuparrow%20%7D%7B116%7D%29%7D%7D)
Really? its .1000 it goes like this "1" is the tenths the "3" is the hundredths the "2" the thousands "5" ten-thousands<span />
