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kicyunya [14]
3 years ago
6

Use the table of values below to select the correct statement.

Mathematics
1 answer:
quester [9]3 years ago
8 0

Answer:

Linear function with rate of change/growth = 2.5, which agrees with the fourth statement listed in the answers options.

Step-by-step explanation:

Notice that both columns of reported x and y values are increasing.

Then examine how the given x-values  increase:

-2, 2, 6, 10, 14  (in steps of 4 units)

and how their corresponding y-values increase:

-2, 8, 18, 28, 38  (in steps of 10 units)

therefore, if we do the rate of change for any pair (x_1,y_1) and (x_2,y_2), we get the following constant rate of change:

\frac{y_2-y_1}{x_2-x_1} = \frac{10}{4} =2.5

Given that this relationship is valid for any pair of (x,y) values. we conclude that the rate of increase is constant, and therefore we are in the presence of a linear function, whose rate of change is 2.5

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8. Identify the common denominator; express each fraction using that denominator; combine the numerators of those rewritten fractions and express the result over the common denominator. Factor out any common factors from numerator and denominator in your result. (It's exactly the same set of instructions that apply for completely numerical fractions.)

9. As with numerical fractions, multiply the numerator by the inverse of the denominator; cancel common factors from numerator and denominator.

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Step-by-step explanation:

For a couple of these, it is helpful to remember that (a-b) = -(b-a).

<h3>8d.</h3>

\dfrac{5}{x+2}+\dfrac{25-x}{x^2-3x-10}=\dfrac{5(x-5)}{(x+2)(x-5)}+\dfrac{25-x}{(x+2)(x-5)}\\\\=\dfrac{5x-25+25-x}{(x+2)(x-5)}=\dfrac{4x}{x^2-3x-10}

___

<h3>9b.</h3>

\displaystyle\frac{\left(\frac{x}{x-2}\right)}{\left(\frac{2x}{2-x}\right)}=\frac{x}{x-2}\cdot\frac{-(x-2)}{2x}=\frac{-x(x-2)}{2x(x-2)}=-\frac{1}{2}

___

<h3>10b.</h3>

\dfrac{3}{x-1}+\dfrac{6}{x^2-3x+2}=2\\\\\dfrac{3(x-2)}{(x-1)(x-2)}+\dfrac{6}{(x-1)(x-2)}=\dfrac{2(x-1)(x-2)}{(x-1)(x-2)}\\\\3x-6+6=2(x^2-3x+2) \qquad\text{multiply by the denominator}\\\\2x^2-9x+4=0 \qquad\text{subtract 3x}\\\\(2x-1)(x-4)=0 \qquad\text{factor; x=1/2, x=4}

Neither solution makes any denominator be zero, so both are good solutions.

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