Answer:
Part 1) 
Part 2) 
Step-by-step explanation:
we know that
The volume of a cube is equal to
where
s is the length of the cube
Part 1)
we have
substitute in the formula
Part 2)
we have
substitute in the formula
Well (3-2^2)5 = -5
and 5-3^2=-4
5-2^2=1
2^2-5=-1
3^2-5=4
but if you have to multiply them by 5 as well then (2^2-5)5=-5
Answer:
George has 8x+2=26 bottles of soap what is x
Step-by-step explanation:
I don’t know but I need the Points sorry for you
Answer:
use logarithms
Step-by-step explanation:
Taking the logarithm of an expression with a variable in the exponent makes the exponent become a coefficient of the logarithm of the base.
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You will note that this approach works well enough for ...
a^(x+3) = b^(x-6) . . . . . . . . . . . variables in the exponents
(x+3)log(a) = (x-6)log(b) . . . . . a linear equation after taking logs
but doesn't do anything to help you solve ...
x +3 = b^(x -6)
There is no algebraic way to solve equations that are a mix of polynomial and exponential functions.
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Some functions have been defined to help in certain situations. For example, the "product log" function (or its inverse) can be used to solve a certain class of equations with variables in the exponent. However, these functions and their use are not normally studied in algebra courses.
In any event, I find a graphing calculator to be an extremely useful tool for solving exponential equations.
