It could be? the 7 could be 7 to literally any number soooo
2 ways: Easy and hard
Hard=A
Easy=B
A: 1/2x+4
work from there so we do fun stuff with it
make something that can be simplified so
1/2x+4 times (2/2)=x+8
now square the whole thing and put the result in a square root thingie
(x+8)^2=x^2+16x+64

multiply the whole thing by 4/4 and put
![\sqrt{16} [\tex] on top so then [tex] \sqrt{x^2+16x+64}](https://tex.z-dn.net/?f=%20%5Csqrt%7B16%7D%20%5B%5Ctex%5D%20on%20top%20so%20then%20%0A%5Btex%5D%20%5Csqrt%7Bx%5E2%2B16x%2B64%7D%20)
times

=

=

to solve it, factor out the 16 in the square root and then square root 16 to get 4
then it will be (4 times square root of equation)/4=square root of equatio
factor square root of equation and square root it and get x+8
divide by 2 to get 1/2x+4
B: 1/2x+4
put stuff that cancels out
1/2x+3x-3x+4+56-56
move them around
3 and 1/2x-3x+60-56
or
2x-3x+1 and 1/2x+30-20+30-36
then just add like terms to solve
Answer:
14 b^5
Step-by-step explanation:
you can do all mathmatical operations to same variables with same power
(9^x) - 3 = 2*3^x
(9^x) - 3 - (2*3^x) = (2*3^x) - (2*3^x)
(9^x) - (2*3^x) - 3 = 0
(3^2)^x - 2*(3^x) - 3 = 0
3^(2x) - 2*(3^x) - 3 = 0
3^(x*2) - 2*(3^x) - 3 = 0
(3^x)^2 - 2*(3^x) - 3 = 0
z^2 - 2*z - 3 = 0 ............ let z = 3^x
(z - 3)(z + 1) = 0
If z-3 = 0, then z = 3 when we isolate z
If z = 3, and z = 3^x, then
z = 3
3^x = 3
3^x = 3^1
x = 1
which is a solutin in terms of x
If z+1 = 0 then z = -1
If z = -1 and z = 3^x, then there are NO solutions for this part of the equation
The quantity 3^x is never negative no matter what the x value is
---------------------------------------------------------------
Answer: x = 1
<h2>10/5=2 test per week</h2><h2>7x2=14</h2><h2>Josh took 14 test in 7 weeks</h2>