My daughter hasnt been taught to do division with a 3 number divisor is their another way to calculate tgis problem?

Mathematics

1 answer:

katovenus [111]3 years ago

6 0

Would long division work?

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In a first -aid kit the ratio of large bandages is 3 to 2. Based on this ratio, how many large bandages are in the kit if there

11Alexandr11 [23.1K]

Answer:

Step-by-step explanation:

The total number of ratio units is 3+2 = 5, and the 3 ratio units representing large bandages make up 3/5 of that total. Thus, large bandages will make up 3/5 of the total number of bandages:

3/5×80 = 48 . . . . number of large bandages in the kit

7 0

4 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:

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