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lesantik [10]
3 years ago
14

Heelp me with this plz im trying to help one of my older sisters

Mathematics
1 answer:
morpeh [17]3 years ago
3 0

Answer:

b

Step-by-step explanation:

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A population increased 13% up to 2,877. What was the population before the increase?
solmaris [256]
113%  is equivalent to 2877
100% will be equivalent to;
\frac{100*2877}{113}
=2547
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The blue and orange lines represent a system. Use thesliders to manipulate the orange line to determinewhich guations would crea
VLD [36.1K]

Answer:

<em>2x + y = -5 </em>

<em> </em>

<em>-2x = y </em>

<em> </em>

<em>2x = 4 -y</em>

Step-by-step explanation:

Use form y = mx + b, and from the lines you will find the following answers equations that have no solution.

7 0
3 years ago
=&gt; Please Help Me With My Math &lt;=
GuDViN [60]

Answer:

B

Step-by-step explanation:

6 0
3 years ago
What are the x-intercepts of the graph of the function f(x)= x^2+4x-12
levacccp [35]
(2,0)
(-6,0)
I just inserted the equation into a graphing calculator. Kinda lazy tonight :/
7 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.

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

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
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