Answer:
x(x+3)(x−10)
Step-by-step explanation:
Hope this was helpful let me know if it was!
Elena should put a point where the two circles intersect and draw line segments connecting that point to points A and B to finish her triangle.
Here is the answer to your problem:
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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.
Answer:
125/2 or 62.5 or 62 1/2 (x=)
Step-by-step explanation:
25/4=x/10 multiply everything by 20
25*20/4=20x/10 simplify
25*5=2x simplify
125=2x divide on both sides
125/2 or 62.5 or 62 1/2=x