What is the scale factor of triangle ABC to triangle DEF. A: 2 B: 1/2 C: 1/6 D: 6

Mathematics

2 answers:

Fudgin [204]3 years ago

8 0

Little tip ** if u r going from a larger shape to a smaller shape, the scale factor is going to be a proper fraction....a number less then 1 but greater then 0

BC = 66......EF = 11.......66 times what is 11
66x = 11
x = 11/66 reduces to 1/6
so 66 * 1/6 = 11

now compare the other sides...
42 * 1/6 = 7.......looks like AC and DF
72 * 1/6 = 12...looks like Ab and DE

so ur multiplying every number in the big triangle by 1/6 , and getting the measures of the little triangle....so ur scale factor is 1/6 <==

BigorU [14]3 years ago

7 0

The answer for this is C

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il63 [147K]

I got -2(2x-3) but it should be C.-4x-2

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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:

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