I hope this helps you
a^6n-3n
a^3n
Answer:
We have the equation:
(ax^2 + 3x + 2b) - (5x^2+bx-3c)= 3x^2 - 9
First, move all to the left side.
(ax^2 + 3x + 2b) - (5x^2+bx-3c) - 3x^2 + 9 = 0
Now let's group togheter terms with the same power of x.
(a - 5 - 3)*x^2 + (3 - b)*x + (2b + 3c + 9) = 0.
This must be zero for all the values of x, then the things inside each parenthesis must be zero.
1)
a - 5 - 3 = 0
a = 3 + 5 = 8.
2)
3 - b = 0
b = 3.
3)
2b + 3c + 9 = 0
2*3 + 3c + 9 = 0
3c = -6 - 9 = -15
c = -15/3 = -5
Then we have:
a = 8, b = 3, c = -5
a + b + c = 8 + 3 - 5 = 6

=

Multiply both sides by 3
z + 6 =

Multiply both sides by 4
(4)z + 6 = (3)2z Simplify
4z + 24 = 6z Subtract 4z from both sides
24 = 2z Divide both sides by 2
12 = z Flip the sides to make it easier to read
z = 12
Answer:
the question is incomplete, so I looked for a similar one:
<em>After the release of radioactive material into the atmosphere from a nuclear power plant, the hay was contaminated by iodine 131 ( half-life, 8 days). If it is all right to feed the hay to cows when 10% of the iodine 131 remains, how long did the farmers need to wait to use this hay? </em>
iodine's half life (we are given x, we need to find b):
0.5A₀ = A₀eᵇˣ
x = 8 days
we eliminate A₀ from both sides
0.5 = eᵇ⁸
ln 0.5 = ln eᵇ⁸
-0.69315 = b8
b = -0.69315 / 8 = -0.08664
since the farmers need to wait until only 10% of the iodine remains (we already calculated b, now we need to find x):
0.1A₀ = A₀eᵇˣ
0.1 = eᵇˣ
ln 0.1 = bx
where b = -0.08664
x = ln 0.1 / -0.08664 = -2.302585 / -0.08664 = 26.58 days