What is the answer to the question?

Mathematics

2 answers:

Verizon [17]3 years ago

6 0

Answer:

5 exterior angles.

Step-by-step explanation: since they all have to add to 360 you can divide 360/5 = 72. You can also check by adding one interior angle plus 72 and checking if you get 180.

kotegsom [21]3 years ago

3 0

Answer:

What is the sum of the exterior angles in a regular 15-gon?

The sum of the exterior angles in any convex polygon, including a regular 15-gon, is 360∘.

Step-by-step explanation:

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Is y= -6x proportional?

Fed [463]

Answer:

yes

Step-by-step explanation:

6 0

3 years ago

Express this to single logarithm

zzz [600]

Answer: $\log_{2}\left(\frac{q^2\sqrt{m}}{n^3}\right)$

We have something in the form log(x/y) where x = q^2*sqrt(m) and y = n^3. The log is base 2.

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Explanation:

It seems strange how the first two logs you wrote are base 2, but the third one is not. I'll assume that you meant to say it's also base 2. Because base 2 is fundamental to computing, logs of this nature are often referred to as binary logarithms.

I'm going to use these three log rules, which apply to any base.

log(A) + log(B) = log(A*B)
log(A) - log(B) = log(A/B)
B*log(A) = log(A^B)

From there, we can then say the following:

$\frac{1}{2}\log_{2}\left(m\right)-3\log_{2}\left(n\right)+2\log_{2}\left(q\right)\\\\\log_{2}\left(m^{1/2}\right)-\log_{2}\left(n^3\right)+\log_{2}\left(q^2\right) \ \text{ .... use log rule 3}\\\\\log_{2}\left(\sqrt{m}\right)+\log_{2}\left(q^2\right)-\log_{2}\left(n^3\right)\\\\\log_{2}\left(\sqrt{m}*q^2\right)-\log_{2}\left(n^3\right) \ \text{ .... use log rule 1}\\\\\log_{2}\left(\frac{q^2\sqrt{m}}{n^3}\right) \ \text{ .... use log rule 2}$