9. $12,50
10. $17.50 (ripoff lol)
11. Clearly Alastair, they got some strong muscles
12. Taryn would be charging $15 an hour and alastair $24.50 an hour. $735 divided by $15 an hour equals 49 hours for Taryn. $735 divided by $24.50 an hour equals 30 hours for Alastair.
13a. 5 minutes to make a sculpture and 7 balloons in a sculpture
13b. Tomas uses 1.4 balloons a minute
13c. 14 balloons (tip. when multiplying a decimal by 10, 100, etc, just move the decimal for the number of zeroes.) 1.4x10 = 14. Simply move the decimal to the right one because there is one zero in 10.
13c blanks. The first blank is 1.4 and the second blank is 14
14. 8 pound bag of dog food is a way better deal!
15. $10.50
16a. The price for 25 whistles is $0.85 each, 50 whistles for $0.72 each, or 80 whistles for $0.75 each. The difference between the highest number and lowest number is $0.13 or 13 cents.
16b. $0.34 or 34 cents
16c. She should order 80 Kazoos <em>not a meme reference</em>
I hope all of this helped! It took forever!
The equation x° = 180° - (37°+ 53°) can be used to find the value of x.
Step-by-step explanation:
Step 1; There are three lines in the given diagram. There are a baseline and two other lines. Out of the other two lines, one extends above and below the baseline whereas the other extends only above. Since the baseline is horizontal and the others are at angles we have the sum of all the three angles as 180° i.e. 37°, x°, and 53°.
37° + x° + 53° = 180°.
Step 2; To solve the value of x, we keep the unknown value at the left-hand side whereas all the known values are taken to the right side of the equation.
x° = 180° - (37°+ 53°).
So the fifth option can be used to determine the value of x.
Answer:
15
Step-by-step explanation:
15 is 5 * 3 so is not prime.
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.
