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barxatty [35]
3 years ago
9

2. An equation is shown below: 5(2x - 3) = 5 Part A: How many solutions does this equation have? (4 points) Part B: What are the

solutions to this equation? Show your work. (6 points)
Mathematics
1 answer:
Minchanka [31]3 years ago
6 0
Distribute first then solve  5(2x-3)  = 10x -15

Thus 10x -15 = 5  now solve for x by adding 15 to both sides of the equation
10x -15 +15 = 5 +15
thus
10x = 20
now divide both sides by 10

\frac{10x}{10}  =  \frac{20}{10}

Thus x = 2
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Simplify the following expression.
erica [24]

Answer:

B)  53x + 74

Step-by-step explanation:

(59x + 64) - (6x - 10) = 59x + 64 - 6x + 10

                                = 53x + 74

3 0
3 years ago
Mr. and Mrs. Simpson went to the movies. They bought some popcorn for $25.50, two boxes of candy for $18 and two drinks for $14.
Ne4ueva [31]
You add all of those numbers together and get your answer which is 60, so basically like this

25.50 + 14.25 + 18 + 2.25 = 60
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2 years ago
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Harrizon [31]

Answer:

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Step-by-step explanation:

7 0
2 years ago
List three exampls of a rate
Tamiku [17]

Answer:

Step-by-step explanation:

It is a special type of ratio

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6 0
2 years ago
Consider the following set of vectors. v1 = 0 0 0 1 , v2 = 0 0 3 1 , v3 = 0 4 3 1 , v4 = 8 4 3 1 Let v1, v2, v3, and v4 be (colu
Alona [7]

Answer:

We have the equation

c_1\left[\begin{array}{c}0\\0\\0\\1\end{array}\right] +c_2\left[\begin{array}{c}0\\0\\3\\1\end{array}\right] +c_3\left[\begin{array}{c}0\\4\\3\\1\end{array}\right] +c_4\left[\begin{array}{c}8\\4\\3\\1\end{array}\right] =\left[\begin{array}{c}0\\0\\0\\0\end{array}\right]

Then, the augmented matrix of the system is

\left[\begin{array}{cccc}0&0&0&8\\0&0&4&4\\0&3&3&3\\1&1&1&1\end{array}\right]

We exchange rows 1 and 4 and rows 2 and 3 and obtain the matrix:

\left[\begin{array}{cccc}1&1&1&1\\0&3&3&3\\0&0&4&4\\0&0&0&8\end{array}\right]

This matrix is in echelon form. Then, now we apply backward substitution:

1.

8c_4=0\\c_4=0

2.

4c_3+4c_4=0\\4c_3+4*0=0\\c_3=0

3.

3c_2+3c_3+3c_4=0\\3c_2+3*0+3*0=0\\c_2=0

4.

c_1+c_2+c_3+c_4=0\\c_1+0+0+0=0\\c_1=0

Then the system has unique solution that is (c_1,c_2c_3,c_4)=(0,0,0,0) and this imply that the vectors v_1,v_2,v_3,v_4 are linear independent.

4 0
2 years ago
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