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Artyom0805 [142]
3 years ago
6

What is the product? Assume x ≠ 0. 2x^2/3{6/x} 3x 4x 3x2 4x2

Mathematics
2 answers:
Serjik [45]3 years ago
7 0

Answer:

4x

Step-by-step explanation:

The given product is;

\frac{2x^2}{3}(\frac{6}{x})

This is the same as;

\frac{2x^2\times 6}{3\times x}

We cancel out the common factors to obtain;

\frac{2x\times 2}{1\times 1}

Multiply out to obtain;

\frac{4x}{1}

Simplify

4x

The second option is the correct choice.

svlad2 [7]3 years ago
7 0

Answer:

4x

Step-by-step explanation:

\frac{2x^2}{3}(\frac{6}{x})

2x^2 can be written as 2*x*x

so the given expression becomes

\frac{2*x*x}{3}(\frac{6}{x})

we cancel out x at the top and bottom

we know 6 divide by 3 is 2

\frac{2*x}{1}(\frac{2}{1})

Now we multiply the numerator with numerator and denominator with denominator

4x

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3 years ago
Select the correct first step to solve the follow equation: −4(x−5)=−7x+10
Otrada [13]

Answer:

x = -10/3

Step-by-step explanation:

−4(x−5)=−7x+10

Use the distributive property to multiply −4 by x−5.

−4x+20=−7x+10

Add 7x to both sides.

−4x+20+7x=10

Combine −4x and 7x to get 3x.

3x+20=10

Subtract 20 from 10 to get −10.

3x=−10

Divide both sides by 3.

x = -10/3

6 0
3 years ago
A line passes through the points (1, 4) and (3, –4). Which is the equation of the line? y = negative 4 x + 8 y = –2x + 6 y = neg
Kruka [31]

Answer:

y = -4x + 8

Step-by-step explanation:

The equation of a line is written in slope-intercept form : y = mx+b

m is the slope

We are given two points, so let's find the slope of the line first:

Slope = ΔY/ΔX = 4 - (-4) / 1 - 3 = 8 / -2 = -4

The slope is -4

So far, our equation is y = -4x + b

We can input a point's x and y value to find b, the y-intercept

Let's use point (1, 4)

4 = -4(1) + b

4 = -4 + b

b = 4 + 4

b = 8

The equation is y = -4x + 8

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Answer

The answer and procedures of the exercise are attached in the following archives.

Explanation  

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

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