[(5^2)^2]^2how to do this

Mathematics

2 answers:

pshichka [43]3 years ago

6 0

Really simple 5^2^2 is equal to 5^4 because a power raised to a power multiple
Your answer is 5^8

Lana71 [14]3 years ago

4 0

Step-by-step explanation:

{(5^2)^2}^2
(5^4)^2
5^8

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Can someone explain why A is the right answer to this question? I'm unsure of how to work out this problem.

joja [24]

So firstly, we have to find the radius of the circular garden before finding the circumference (the amount of fencing needed to surround the garden). To find the radius, use the area formula ( $A=\pi r^2$ ), plug in the area of the garden (36 ft^2) and solve for r as such:

$36=\pi r^2\\ \frac{36}{\pi}=r^2\\ \frac{\sqrt{36}}{\sqrt{\pi}}=r\\ \frac{6}{\sqrt{\pi}}=r$

So that we know the radius, plug that into the circumference equation ( $C=2\pi r$ ) to solve:

$C=2\pi*\frac{6}{\sqrt{\pi}}\\ C=\frac{12\pi}{\sqrt{\pi}}\\ C=12\sqrt{\pi}$

Your answer is A. 12√π.

4 0

2 years ago

Please need help with this one

Sergio039 [100]

Answer:

$7 \times 1 > \frac{5}{9} simplify \\$

if you don't its still right though

Step-by-step explanation:

only if you don't simplyfy if you do then just subtract and you will get a another answer which both are right

4 0

2 years ago

Read 2 more answers

Enter the explicit rule for the sequence

Softa [21]

Let's take a look at the first few numbers in the sequence based on the given rule:

$a_{1}=6\\ a_{2}= \frac{1}{4}(6)\\ a_3= \frac{1}{4}\big( \frac{1}{4}\big)6= \big(\frac{1}{4}\big)^2(6)\\ a_{4}= \frac{1}{4} \big(\frac{1}{4}\big)^26= \big(\frac{1}{4}\big)^3(6)$

Inspecting this pattern it seems like the power $\frac{1}{4}$ is being raised to is always one less than the number of the sequence, so if we were on the nth number in the sequence, that part of the expression would be $\big(\frac{1}{4} \big)^{n-1}$ . We also know that we'll be multiplying whatever we get from that by 6, so we can write the full explicit rule for our sequence as

$a_n=6\big( \frac{1}{4} \big)^{n-1}$

Where $a_n$ is the nth number in our sequence.