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wlad13 [49]
3 years ago
6

in a proportional relationship the ratio y/x is constant. Show that this ratio is not constant for the equation y=a-14

Mathematics
1 answer:
Colt1911 [192]3 years ago
8 0

To show that the ratio \frac{y}{x} is not constant for the function y=a-14 we only need to evaluate this function at 2 points and then show that the ratio is not the same for those two points. Note that here a represents x.  

To find the counter examples here, I will use the values a=20  and a=28. For a = 20 , the equation tells us that the output is

y=20-14=6. For this value of a the point is point is (20,6). The ratio \frac{y}{x} =\frac{6}{20}. For the a=28, y=28-14=14. For this value of a the  point is (28,14) . The ratio \frac{y}{x} =\frac{14}{28} =\frac{1}{2}. Clearly this ratio is not constant for all ordered pairs.

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3 years ago
5. When y varies directly with x, the value of y = -2 when x = 4. What is the value of y when x = 8?
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Answer:

-4

Step-by-step explanation:

if y = -2 then x = 4

what is y when x = 8

\frac{ - 2}{x}  =  \frac{4}{8}  \\  - 16 = 4x \\ x =  - 4

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Find the horizontal and vertical asymptotes of​ f(x). ​f(x) equals = StartFraction 6 x Over x plus 2 EndFraction 6x x+2 Find the
miss Akunina [59]

Answer:

The horizontal asymptote can be described by the line y = 6

The vertical asymptote can be described by the line x = -2

Step-by-step explanation:

* <em>Lets the meaning of vertical and horizontal asymptotes</em>

- <u><em>Vertical asymptotes</em></u> are vertical lines which correspond to the zeroes

  of the denominator of a rational function

- <u><em>A horizontal asymptote</em></u> is a y-value on a graph which a function

 approaches but does not actually reach

- If the degree of the numerator is less than the degree of the

 denominator, then there is a horizontal asymptote at y = 0

- If the degree of the numerator is greater than the degree of the

 denominator, then there is no horizontal asymptote

- If the degree of the numerator is equal the degree of the denominator,

 then there is a horizontal asymptote at y = leading coefficient of the

 numerator ÷ leading coefficient of the denominator

* <em>Lets solve the problem</em>

∵ f(x)=\frac{6x}{x+2}

∵ The numerator is 6x

∵ The denominator is x + 2

∴ The numerator and the denominator have same degree

∵ The leading coefficient of the numerator is 6

∵ The leading coefficient of the denominator is 1

∴ There is a horizontal asymptote at y = 6/1

∴ <em>The horizontal asymptote can be described by the line y = 6</em>

- Put the denominator equal zero to find its zeroes

∵ The denominator is x + 2

∴ x + 2 = 0

- Subtract 2 from both sides

∴ x = -2

∴ <em>The vertical asymptote can be described by the line x = -2</em>

6 0
3 years ago
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Pepsi [2]
The answer is A. dodecagon.
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