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elena55 [62]
3 years ago
7

Which polygons are congruent? Select

each}" alt="\bold{each}" align="absmiddle" class="latex-formula"> correct answer

Mathematics
1 answer:
gregori [183]3 years ago
8 0
<h3>Answer: Everything but the lower right hand corner</h3>

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

Explanation:

Notice for the corners mentioned, we have the figures with corresponding angles that are the same (shown by similar arc markings) and they have congruent corresponding sides as well (aka they are the same length shown by similar tickmarks). Rotating one figure has it transform into the other.

The only time this does not happen is with the pair of figures in the bottom right hand corner. One square has side lengths of 20, the other has side lengths of 25. The two figures are not congruent due to the side mismatch.

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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.

  1. log(A) + log(B) = log(A*B)
  2. log(A) - log(B) = log(A/B)
  3. 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}

8 0
3 years ago
Find height of triangle given hypotenuse and angle
harkovskaia [24]
Given h is the height, H is the hypotenuse and A is the base angle,
then Sin(A) = h/H
so h = HSin(A)
7 0
4 years ago
Jay Spring pays $350.00 annually for his health insurance. This is 1/3 of the total cost of the premium; his company pays the re
WINSTONCH [101]

Q: How much did Jay have to pay excluding his share of the insurance premium?

A: $1800+$200 = $2000

Q: How much did Jay's company pay for his insurance premium?

A: $700. If Jay's $350 is 1/3 of the premium , then Jay's company pays 2*$350=$700 as rest of his premium.

Q: Jay paid 10% and the plan paid 90% beyond the deductible. How much did Jay's insurance company pay total?

A: Jay's insurance company paid $16200. Given that Jay paid $1800 beyond his deductible of $200 (and that is 10% of the actual cost) means that his plan (insurance company) paid 90%=9*$1800=$16200.

Q: How much did Jay have to pay total, including his share of the premium?

A: Jay paid $2350. He paid $200 deductible + $1800 beyond deductible + $350 premium



8 0
3 years ago
Read 2 more answers
If f and g are functions, then (fog)(x) is always equal to (gof)(x)<br><br> true or false
valentina_108 [34]

Answer:

False

Step-by-step explanation:

Here is a counter example:

● f(x) = x^2

● g(x) = 3x

● (f○g)(x) = (3x)^2 = 9x^2 (1)

● (g○f)(x) = 3(x^2) = 3x^2 (2)

We can see that (1) and (2) aren't equal to each others.

4 0
4 years ago
There are 5 white balls, 8 red balls, 7 yellow balls and 4 green balls in a container. A ball is chosen at random.
Nookie1986 [14]

Answer:

green=4/24    yellow=7/24     black=0/24    red, yellow=15/24

Step-by-step explanation:

7 0
3 years ago
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