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How many different groups of three kids can you make from seven kids?<br> 35<br> 210<br> 6<br> 24

Neko [114]

The answer is 35 different groups.

6 0

2 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

2 years ago

Mark reduce the size of a photo to a height of 4 inches. What is the new width if it was originally 16 inches tall and 8 inches

uranmaximum [27]

Answer:

Step-by-step explanation:

new height is ¼ of original height

new width = ¼×8 = 2 inches

4 0

2 years ago

Read 2 more answers

Lee needed to make drapes the length from the floor to the top of the window. The wall beneath the window had the height of 2 2/

Tatiana [17]

Since drapes have to be covered all over the window and the beneath of the window, both lengths have to be added.

(2 2/3) + ( 2 3/4)
(8/3) + (11/4)
(32+33)/12
65/12
5 5/12

7 0

3 years ago

Show work and Factor 4x^2+xy-18y^2

grigory [225]

Answer:

$4x^2+xy-18y^2=(4x+9y)\,(x-2y)$

Step-by-step explanation:

Let's examine the following general product of two binomials with variables x and y in different terms:

$(ax+by)\,(cx+dy)= ac\,x^2+adxy+bcxy+bdy^2$

so we want the following to happen:

$a\,c = 4\\ad+bc=1\\bd--18$

Notice as well that $ad+bc =1$ means that those two products must differ in just one unit so, one of them has to be negative, or three of them negative. Given that the product $bd=-18$ , then we can consider the case in which one of this (b or d) is the negative factor. So let's then assume that $a\,\,and \,\,c$ are positive.