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Maths-

General

Easy

Question

Which of the following equations relates y to x for the values in the table above?

$y = \frac{1}{2} \cdot {(\frac{5}{2})}^{x}$
$y = 2 \cdot {(\frac{3}{4})}^{x}$
$y = \frac{3}{4} x + 2$
$y = \frac{7}{2} x - \frac{3}{4}$

hint Hint:

n such question, we check the options by putting values.
Explanation:

We have given different values of (x, y) in the tabular form

We have to find the equation that relates y to x for the values in the table form.

We will put the values of (x, y) in options and then check, which equation satisfy the given values of (x, y).

The correct answer is: $y equals 7 over 2 x minus 3 over 4$

Step 1 of 1:
We have given different values of (x, y)
We first check option (a)
As given in the table for $x equals 1 comma y equals 11 over 3$ ,
So,
$y equals 1 half open parentheses 5 over 2 close parentheses to the power of x$
$11 over 3 equals 1 half open parentheses 5 over 2 close parentheses to the power of 1$
$11 over 3 equals 5 over 4$
Which is a contradiction,
So, Option (A) is incorrect.
For option (B)
$y equals 2 open parentheses 3 over 4 close parentheses to the power of x$
$25 over 4 equals 2 open parentheses 3 over 4 close parentheses to the power of 1$
$25 over 4 equals 6 over 4$
Which is a contradiction
So, Option (B) is incorrect.
Step 2 of 2
For option (c)
$y equals 3 over 4 x plus 2$
$11 over 4 equals 3 over 4 left parenthesis 1 right parenthesis plus 2$
$11 over 4 equals 11 over 4$
$text Now put end text x equals 2$
$y equals 3 over 4 x plus 2$
$25 over 4 equals 3 over 4 left parenthesis 2 right parenthesis plus 2$
$25 over 4 equals 14 over 4$
This is aga in contradiction
So, Option (c) is also incorrect
For option (d)
$y equals 7 over 2 x minus 3 over 4$
$11 over 4 equals 7 over 2 left parenthesis 1 right parenthesis minus 3 over 4$
$11 over 4 equals 11 over 4$
This is true.
So, Option (d) is correct.

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