Find the product -8. (-8)=

Mathematics

2 answers:

My name is Ann [436]2 years ago

7 0

The product of -8. (-8)= 64

dezoksy [38]2 years ago

5 0

The answer is -8.(-8)= 64

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To participate in a competition, each team must follow the rules about the allowable number of adults and children on a team.

Snowcat [4.5K]

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2 years ago

When Javier and his father went to dinner, the bill was $34. If they tipped the waiter 20%, how much did they spend in total?

Kamila [148]

Answer:

Not sure if this is right, but I got $40.8.

Step-by-step explanation:

I first turned 20% into a decimal. (0.20).

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Read 2 more answers

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:

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