Answer:
8) answer=5 9)answer=15
Step-by-step explanation:
all you have to do is replace in the following formula

Where
X1: is the first number of the first point
Y1: is the second number of the first point
X2: is the first number of the second point
Y2: is the second number of the second point
you replace in the formula and solve
I attach solution
Answer: My guess is $4.55
Step-by-step explanation:
2.50 + 1.75 = 4.25 (This is the total of everything you bought)
4.25 x 0.07= 0.2975 (0.2975 is 7% of 4.25 but because this is money I will round it to <em>0.30</em>)
(0.30 is the tax so now add it to the original amount)
$4.25 + $0.30 = $4.55
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.
157.5
$5.75 x 10 = 57.5
$2.25 x 20 = 45
$5.00 x 11 = 55
then you add all them together and you get 157.5