Reducing fractions 32/40

Mathematics

1 answer:

jenyasd209 [6]2 years ago

4 0

32/40 divided by 8/8= 4/5

The answer is 4/5.

Hope this helps!

You might be interested in

a sticker price on a new vehicle is $33985 , but the dealer cost is $27700 and you want to offer 4% over the dealer cost. what w

Alexandra [31]

To find the price of the dealer cost, you should multiply 27700 by 4% and add the amount to 27700.

4%, in decimal form (so we can easily multiply it), is 0.04.

27700 * 0.04 = 1108

27700 + 1108 = 28808

The answer is 28808 dollars.

6 0

2 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.

===========================================================

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:

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