The mixed number have to become a improper fraction and then you have to subtract like this.
so 7 7/8 you have to multiply 7 times 8 and then add it by 7 .
You do the same thing with the other mixed number 3 times 4 plus 1
So the first one is 63/8 - 13/4 =
first the 4 in the 13/8 have to become 8 . so you do 4 times 2 = 8 and multiply the 13 by 2 = 26 .
so equal
63/8 - 26/8 = then you have to do 63 - 26 = 37 so is 37/8 and then simplify or divide .
when you divide you suppose to get 4 5/8
Answer:
it's c
it might be wrong because i dont know the question
but as a guess i pick c
in every multiple choice questton there is always a 60% chance the answer is c
Answer:
Y<u>></u>-x-5
Step-by-step explanation:
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.
