Write a trinomial expression that is equivalent to (2x+5)(3x-2)

1 answer:

8 0

2x times 3x is 6x to the second
2x times -2 is -4x
5 times 3x is 15x
-2 times 5 is -10
6x to the second plus 9x minus 10

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PLEASE HELP 20 POINTS!!!

grigory [225]

The answer is A i hope this help

6 0

3 years ago

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My problem is : P=2L+2W solve for L

natima [27]

Answer:
L = P/2 - W

Explanation:
Step 1 - Start by factoring out the two

P = 2L + 2W
P = 2(L + W)

Step 2 - Divide both sides of the equation by two

P = 2(L + W)
P/ 2 = 2(L + W)/ 2
P/ 2 = L + W

Step 3 - Subtract W from both sides of the equation

P/ 2 = L + W
P/ 2 - W = L + W - W
P/ 2 - W = L

3 0

3 years ago

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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.

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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:

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