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erastova [34]
3 years ago
7

Which statement best explains if the graph correctly represents the proportional relationship y = 0.5x?

Mathematics
2 answers:
yan [13]3 years ago
7 0
I believe the correct answer from the choices listed above is the first option. The statement that best explains the graph that shown below would be yes, the points shown on the line would be part of y = 0.5x. Hope this answers the question. Have a nice day.
ehidna [41]3 years ago
3 0
The <u>correct answer</u> is:

<span>A) Yes, the points shown on the line would be part of y = 0.5x.

Explanation:

Looking at the graph, the line goes through (0, 0).  If x=0, we have
y=0.5(0)=0.  
This is the y-coordinate of the point, so it goes through this.

The equation for this line is in slope-intercept form, y=mx+b, where m is the slope and b is the y-intercept.  In this equation, m=0.5 and b=0.  This means the slope is 0.5 or 1/2; since slope is rise/run, this means the line rises 1 for every horizontal increase of 2.  

This means starting from (0, 0), we would go up 1 and over 2, creating the point (2, 1).  Looking at the graph, the line goes through this point.

Next we would go up 1 and over 2 again, to (4, 2).  Again, the line goes through this point.

We could follow all of the points on this line and they all fit.</span>
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Write an equation for the horizontal line that contains point E(-3, -1)
ryzh [129]
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What is the approximate solution to this equation? 15(3)^2x=90
mel-nik [20]

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x ≈ 0.67

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Help solve for number 6 please !!<br> :)
Semenov [28]

Answer:

x: {-3, 5}

Step-by-step explanation:

Step 1. Find least common denominator

Step 2. Multiply missing factors on top AND bottom

Step 3. Combine like terms

Step 4. Simplify (get rid of denominator)

Step 5. Solve for x

\frac{x}{x-5} +\frac{3}{x+2} =\frac{7x}{x^2-3x-10}            LCD = x^2-3x-10   OR  (x-5)(x+2)

\frac{x}{x-5}(\frac{x+2}{x+2})  +\frac{3}{x+2}(\frac{x-5}{x-5})  =\frac{7x}{x^2-3x-10}

\frac{x(x+2)}{(x-5)(x+2)} +\frac{3(x-5)}{(x+2)(x-5)} =\frac{7x}{(x-5)(x+2)}

\frac{x(x+2)+3(x-5)}{(x-5)(x+2)} =\frac{7x}{(x-5)(x+2)}

\frac{x^2+2x+3x-15}{(x-5)(x+2)} =\frac{7x}{(x-5)(x+2)}

\frac{x^2+5x-15}{(x-5)(x+2)} =\frac{7x}{(x-5)(x+2)}

x^2+5x-15=7x

x^2-2x-15

(x-5)(x+3)

<u>x = -3, 5</u>

8 0
3 years ago
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