Best Answer:
<span>
to multiply fractions you simply take the denominator of both fractions
and multiply them, do the same to the numerator, and you have your
answer.
ex.
2 5 10
-- * -- = ----
4 8 32
then, of course, you would need to simplify.
so the answer is
10(/2) 5
-------- = -----
32(/2) 16
To multiply mixed numbers, you need to convert them to improper
fractions, multiply and then convert them back to a mixed number. For
example:
2 1/2 x 5 1/3 = 5/2 x 16 /3
numerator 5 x 16 = 80 denominator 2 x 3 = 6
so now you convert 80/6 into a mixed number
80 / 6 = 13 and 2/6 or, reduced, 13 1/3. </span>
<h3>
Answer:</h3>
- C. (9x -1)(x +4) = 9x² +35x -4
- B. 480
- A. P(t) = 4(1.019)^t
Step-by-step explanation:
1. See the attachment for the filled-in diagram. Adding the contents of the figure gives the sum at the bottom, matching selection C.
2. If we let "d" represent the length of the second volyage, then the total length of the two voyages is ...
... (d+43) + d = 1003
... 2d = 960 . . . . . . . subtract 43
... d = 480 . . . . . . . . divide by 2
The second voyage lasted 480 days.
3. 1.9% - 1.9/100 = 0.019. Adding this fraction to the original means the original is multiplied by 1 +0.019 = 1.019. Doing this multiplication each year for t years means the multiplier is (1.019)^t.
Since the starting value (in 1975) is 4 (billion), the population t years after that is ...
... P(t) = 4(1.019)^t
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.
-x is 13 units to the right of 0 or -13 units to the left of 0.
Hope this helps :)
Step-by-step explanation:
A(-5,1) and B(0,4)
X1=0 , Y1=4 X2= -5 , Y2=1
y - 4 = ⅗(x-0) ==> y - 4 = ⅗ X ==> y = ⅗ X + 4
<u>C</u> is the answer