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Alina [70]
2 years ago
5

Combining !12x + 5x +4-16​

Mathematics
2 answers:
allochka39001 [22]2 years ago
5 0

Answer:

17x-12

Step-by-step explanation:

combine like terms. 12x and 5x go together to make 17x. 4 and -16 go together to make -12

ololo11 [35]2 years ago
5 0
Combine like terms, 12x+5x=17x
and 4-16= -12
17x-12
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2/3 (x-7) = -2 please please help!!
Maslowich

Answer:

x = 4.

Step-by-step explanation:

Distribute, isolate and solve.

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2 years ago
The primary purpose of Passage 1 is to
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Correct answer is E. The primary purpose of Passage 1 is to Highlight a concern.

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6 0
1 year ago
Find the number that is 42 less than twice the<br> opposite of itself.
maria [59]
42 - -84=-42

Hope this helps :)
8 0
2 years ago
If a data value in a normal distribution has a negative z-score, which of the following must be true?
mel-nik [20]
Correct Answer:
The data value must be less than the mean.

The formula for the z value is:

z-value = (Data value - Mean) / Standard Deviation

The z-value will be negative when the Numerator of the above expression will be negative. The numerator will be negative only when the Data Value is less than the mean value. For example, for a data value of 5 and mean value 8, the numerator will be -3, and when it will be divided by standard deviation, the resulting z value will be negative.

So, Third option is the correct answer. 
4 0
3 years ago
Read 2 more answers
Suppose you choose a team of two people from a group of n &gt; 1 people, and your opponent does the same (your choices are allow
jonny [76]

Answer:

The number of possible choices of my team and the opponents team is

 \left\begin{array}{ccc}n-1\\E\\n=1\end{array}\right     i^{3}

Step-by-step explanation:

selecting the first team from n people we have \left(\begin{array}{ccc}n\\1\\\end{array}\right)  = n possibility and choosing second team from the rest of n-1 people we have \left(\begin{array}{ccc}n-1\\1\\\end{array}\right) = n-1

As { A, B} = {B , A}

Therefore, the total possibility is \frac{n(n-1)}{2}

Since our choices are allowed to overlap, the second team is \frac{n(n-1)}{2}

Possibility of choosing both teams will be

\frac{n(n-1)}{2}  *  \frac{n(n-1)}{2}  \\\\= [\frac{n(n-1)}{2}] ^{2}

We now have the formula

1³ + 2³ + ........... + n³ =[\frac{n(n+1)}{2}] ^{2}

1³ + 2³ + ............ + (n-1)³ = [x^{2} \frac{n(n-1)}{2}] ^{2}

=\left[\begin{array}{ccc}n-1\\E\\i=1\end{array}\right] =   [\frac{n(n-1)}{2}]^{3}

4 0
3 years ago
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