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Aleksandr-060686 [28]
3 years ago
11

What an equivalent expression for 48 + 18x

Mathematics
2 answers:
FromTheMoon [43]3 years ago
7 0

Answer:

6(8+3x)

Step-by-step explanation:

48+18x

We can factor out an 6 from each term in the expression

6(8+3x)

White raven [17]3 years ago
7 0

Answer:

  6(8 + 3x)

Step-by-step explanation:

There are an infinite number of equivalent expressions. We presume you have a list to choose from.

One of the more useful ones is the expression you get when you factor out the greatest common factor of the two terms. That factor is 6, so ...

  48 + 18x = 6(8 + 3x)

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Students roll a 6-sided number cube and spin the shown spinner. blue blue orange red orange red orange purple What is the probab
Ede4ka [16]

Answer:

1. 8.333% 2. I don't know ( sorry )

6 0
2 years ago
Find the zeros of polynomial function and solve polynomials equations.<br> f(x)=24x^3-64x^2-21x+56
8090 [49]

Since this polynomial has 4 terms, factoring by grouping should be the first thing we try here.

So, we have:

8x^2(3x-8)-7(3x-8)=0 \implies\\ (8x^2-7)(3x-8)=0

So, we can use ZPP to find out roots:

3x-8 =0 \implies\\ x=\frac{8}{3}\\ 8x^2-7=0 \implies \\ 8x^2=7 \implies \\ x^2=\frac{7}{8} \implies \\ x=\pm\frac{\sqrt{7}}{\sqrt{{8}}}\\ \text{Rationalizing denominator:}\\ x=\pm\frac{\sqrt{56}}{8}

So our three roots are:

x \in \{\frac{8}{3},\frac{\sqrt{56}}{8},-\frac{\sqrt{56}}{8}\}

8 0
2 years ago
Evaluate 3 to the second power + (8 - 2 ) • 4 - 6/3
puteri [66]

Answer:

31

Step-by-step explanation

(3^2) + ((8-2)x4)-(6/3)

9+(24)-2

31

7 0
2 years ago
Solve for the roots in simplest form using the quadratic formula:<br> 4x^2 -28x= -53
irinina [24]

Answer:

x = (28 +- root -64) / (8)

Step-by-step explanation:

4x^2 -28x= -53 = 0

x = [-(-28)+- root (-28)^2 - 4 * 4 * 53] / 2 *4

x = (28 +- root 784 - 848) / (8)

x = (28 +- root -64) / (8)

3 0
3 years ago
Which row operation will triangularize this matrix?
Ludmilka [50]
Triangularizing matrix gives the matrix that has only zeroes above or below the main diagonal. To find which option is correct we need to calculate all of them.
In all these options we calculate result and write it into row that is first mentioned:

A)R1-R3
\left[\begin{array}{ccc}-1&0&0|0\\0&1&1|6\\2&0&1|1\end{array}\right]

B)2R2-R3
\left[\begin{array}{ccc}1&0&1|1\\-2&2&1|4\\2&0&1|1\end{array}\right]

C)-2R1+R3
\left[\begin{array}{ccc}0&0&-1|-1\\0&1&1|6\\2&0&1|1\end{array}\right]

D)2R1+R3
\left[\begin{array}{ccc}4&0&3|3\\0&1&1|6\\2&0&1|1\end{array}\right]

E)3R1+R3
\left[\begin{array}{ccc}5&0&4|4\\0&1&1|6\\2&0&1|1\end{array}\right]

None of the options will triangularize this matrix. The only way to <span>triangularize this matrix is
R3-2R1
</span>\left[\begin{array}{ccc}1&0&1|1\\0&1&1|6\\0&0&-1|-1\end{array}\right]
<span>
This equation is similar to C) but in reverse order. Order in which rows are written is important.</span>
4 0
3 years ago
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