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3 years ago

Simplify the expression so there is only one positive power for each base. (5^-2 x 4^-4)^-2

Mathematics

2 answers:

aksik [14]3 years ago

8 0

I believe you can cancel the ^-2 and ^-4

giving you 5 x 4^2

Which is 5 * 16

Total of 80

Oliga [24]3 years ago

6 0

40960000 is the answer to the equation

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Help me please and thankyou

klemol [59]

Answer:

$\boxed{Option B}$

Step-by-step explanation:

=> $(-5*4)^3$

Using the distributive property

=> $(-5)^3*(4)^3$

5 0

4 years ago

Read 2 more answers

If you deposit $300 in an account with a 6% interest rate, how much will be in your account after 1 year?

Ivahew [28]

You would have $318. Hope this helps out!

5 0

3 years ago

In a pile of coins there are 7 more quarters than nickels. If there is a total of$2.65 in coins, how many quarters are there?

Alinara [238K]

A quarter is equivalent to 25 cents or $0.25 and a nickel is equal to 5 cents or $0.05. In the given in this item, if there are x nickels then, there are x + 7 quarters. With these and the total amount known, we can now compute for the value of x.
(x)(0.05) + (x + 7)(0.25) = 2.65
Simplifying,
0.05x + 0.25x + 1.75 = 2.65
Simplifying further,
0.3x = 0.9
The value of x from the equation is 3. Hence, there are 10 quarters.

6 0

3 years ago

Read 2 more answers

Santa Claus has been experiencing some recent back pain and is unable to carry large bags of toys over his shoulders. As a resul

jeka57 [31]

Answer:

31 pounds

Step-by-step explanation:

From the information given, you can write the following formula to calculate the total cost of sending gifts:

Total cost of sending gifts=2.50+0.85x, where x is the weight of the gifts in pounds

Also, you know that the total cost of sending gifts is $28.85 and you can replace this value and solve for x:

28.85=2.50+0.85x

28.85-2.50=0.85x

26.35=0.85x

x=26.35/0.85

x=31

According to this, the answer is that the gifts weighed 31 pounds.

6 0

3 years ago

Probability extension: Jerry did only 5 problems of one assignment. What is the probability that the problems he did comprised t

zimovet [89]

Answer:

(b) 1/792

Step-by-step explanation:

The complete question is;

Counting: Grading One professor grades homework by randomly choosing 5 out of 12 homework problems to grade.

(a) How many different groups of 5 problems can be chosen from the 12

problems?

(b)Probability extension: Jerry did only 5 problems of one assignment. What is the probability that the problems he did comprised the group that was selected to be graded?

In (a)

Apply the formula

$\frac{n!}{(n-r)!(r!)}$

where n=12 and r=5

substitute values

$=\frac{12!}{(12-5)!(5!)} \\\\\\=\frac{12!}{(7!)(5!)} \\\\\\=\frac{12*11*10*9*8}{5*4*3*2*1} \\\\\\=\frac{95040}{120} \\\\\\=792$

In (b)

If Jerry did only 5 problems of one assignment then the probability will be

$\frac{5}{12} *\frac{4}{11} *\frac{3}{10} *\frac{2}{9} *\frac{1}{8} =\frac{1}{792}$

4 0

3 years ago