Answer:
1205.75999
Step-by-step explanation:
To do this you first have to set this up vertically.
First you multiply through with the 2 (44×2=88). Then next, you have to multiply through with the 1 (44×1=44) But since this is the second line, you have to add the 0 to the end (440).
Now you simply add 88+440=528.
Answer:
186 °F
Step-by-step explanation:
The difference is found by subtracting the smaller from the larger:
136 -(-50) = 136 +50 = 186
The difference between the two temperatures is 186 °F.
Answer:
80 square meters
Step-by-step explanation:
A rectangle that is partitioned into two rectangles; rectangle A and rectangle B
Rectangle A:
Top = 5 meters
Side = 8 meters
Area of rectangle A = length × width
= 5 meters × 8 meters
= 40 meters ²
Rectangle B:
Top = 5 meters
Side = 8 meters
Area of rectangle B = length × width
= 5 meters × 8 meters
= 40 meters ²
Total area of the partitioned rectangle = area of rectangle A + area of rectangle B
= 40 meters ² + 40 meters ²
= 80 meters ²
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.
__
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.
__
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.
