Answer:
7
Step-by-step explanation:
1/2(x)=7/2
divide both sides by 1/2
x=
÷![\frac{1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D)
x=
·![\frac{2}{1}](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B1%7D)
x=14/2=7
Do recall that squaring and the *radical sign* cancel each other out... like so:(
![\sqrt{a}](https://tex.z-dn.net/?f=%20%5Csqrt%7Ba%7D%20)
)
![^{2}](https://tex.z-dn.net/?f=%20%5E%7B2%7D%20)
= a
When you put it that way, it isn't enough :P
(
![\sqrt{a}](https://tex.z-dn.net/?f=%20%5Csqrt%7Ba%7D%20)
)
![^{2}](https://tex.z-dn.net/?f=%20%5E%7B2%7D%20)
= a
(
![\sqrt{8x+1}](https://tex.z-dn.net/?f=%20%5Csqrt%7B8x%2B1%7D%20)
)
![^{2}](https://tex.z-dn.net/?f=%20%5E%7B2%7D%20)
=?
so you start with
(
![\sqrt{8x+1}](https://tex.z-dn.net/?f=%20%5Csqrt%7B8x%2B1%7D%20)
)
![^{2}](https://tex.z-dn.net/?f=%20%5E%7B2%7D%20)
=
![(5)^{2}](https://tex.z-dn.net/?f=%20%285%29%5E%7B2%7D%20)
8x+1=25 <-- subtract 1 to both sides
8x=24 <- divide 8 to both sides
x= 3
To find out if it's an extraneous solution ask yourself: It mustn't result in a radical that I like to call... 'illegal'. Plug it into the radicand 8x+1 and make sure you get something that is not a negative number.... so, DO you get a negative number when you plug in x = 3 into the radicand?
(extraneous solution is a invalid solution)
x=3 not extraneous
All you need to know is that
![\displaystyle \lim_{x\to\infty}a^x=\begin{cases} \infty \text{ if } a>1\\1 \text{ if } a=1\\0 \text{ if } 0\leq a](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bx%5Cto%5Cinfty%7Da%5Ex%3D%5Cbegin%7Bcases%7D%20%5Cinfty%20%5Ctext%7B%20if%20%7D%20a%3E1%5C%5C1%20%5Ctext%7B%20if%20%7D%20a%3D1%5C%5C0%20%5Ctext%7B%20if%20%7D%200%5Cleq%20a%3C1%5Cend%7Bcases%7D)
So, the first function grows exponentially and the second decays exponentially. The rate of growth/decay are the bases of the exponentials, which means 1.08 and 0.205
(0,4) is the y intercept
Hope this helps :)
55+10.00x ≥ 150 x ≥ 9.5 hours I hope that helps