Answer:
Step-by-step explanation:
Let's organise our information :
- the function is (1/2)x²+2x+3
we want to khow the value of x that gives us : (1/2)x²+2x+3=x+7
Now the trick is to write this expression as a quadratic equation with zero at one side :
- (1/2)x²+2x+3-x-7=0
- (1/2)x²+x-4=0
Now let's solve this equation :
Δ=1²-4*(1/2)*(-4)
= 9
So we have two solutions :
So the solutions are -4 and 2
let's first off apply a log rule of cancellation, keeping in mind that, first off is ln(), not in(), and that ln() is just a shortcut to logₑ.
![\bf \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{\stackrel{\textit{we'll use this one}}{\downarrow }}{a^{log_a x}=x} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ e^{ln(x)}\implies e^{log_e(x)}\implies x \\\\\\ \cfrac{d}{dx}\left[ e^{ln(x)} \right]\implies \cfrac{d}{dx}[x]\implies 1](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7BLogarithm%20Cancellation%20Rules%7D%0A%5C%5C%5C%5C%0Alog_a%20a%5Ex%20%3D%20x%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Cstackrel%7B%5Ctextit%7Bwe%27ll%20use%20this%20one%7D%7D%7B%5Cdownarrow%20%7D%7D%7Ba%5E%7Blog_a%20x%7D%3Dx%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0Ae%5E%7Bln%28x%29%7D%5Cimplies%20e%5E%7Blog_e%28x%29%7D%5Cimplies%20x%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7Bd%7D%7Bdx%7D%5Cleft%5B%20e%5E%7Bln%28x%29%7D%20%5Cright%5D%5Cimplies%20%5Ccfrac%7Bd%7D%7Bdx%7D%5Bx%5D%5Cimplies%201)
Answer:
SA = 680 units^2
V =1200 unit^3
Step-by-step explanation:
If the radius is 5 then the diameter is 2r = 2*5 = 10
The height would remain the same
Surface area of a prism is given by
SA = 2 (lw+lh+hw)
l =d = 10 w = d = 10 h = 12
SA = 2 ( 10*10 + 10 * 12+ 12* 10)
SA = 2( 100+120+120)
SA = 2(340)
SA = 680 units^2
V = l*w*h = 10*10*12 =1200 unit^3