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3 years ago

What is the factorization of 3x^2 + 6x - 24 ???

Mathematics

1 answer:

xxTIMURxx [149]3 years ago

4 0

Answer:

3(x-2) (x+4)

Step-by-step explanation:

Factor the polynomial

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Two times what equals four teen

Helen [10]

Answer:

Step-by-step explanation:

Set up an equation where x is the number

2x = 14

Divide by 2 on each side

x = 7

So, the number is 7.

7 0

3 years ago

Read 2 more answers

Which is the simplified form of m^-8p^0

kipiarov [429]

Answer:

1 /m ^8

Step-by-step explanation:

Hope this helps

5 0

3 years ago

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A rectanglular swimming pool is twice as long as it is wide. A small concrete walk was surrounds the pool. The walk way is a con

xxMikexx [17]

<span>subtract 16 from both sides, you will get
12x= 180
divide by 12
x=15 ft.
Answer: width = 15 ft.; length = 30 ft.

</span>

8 0

3 years ago

A rectangle has an area of 221 square units. The width of the rectangle is 13 units. The length of the rectangle is 3x + 2. What

vodka [1.7K]

The length is 17 and x=5

13 times 3x+2 = 221 so solve by dividing 221 by 13= 17.
Now we know that 3x+2= 17 so 3x= 15
x=5

8 0

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:

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

8 0

3 years ago