What Is The Rate Of Change Of The Linear Relationship Modeled In The Table?

what is the rate of alter of the linear relationship modeled in the table?

what is the rate of change of the linear relationship modeled in the tabular array? x y 1 2 3 five 5 eight 7 eleven (four points) a) negative three over two b) ii over three c) 1 d) three over two

Answers

The correct answer is 3/two.

Slope = Charge per unit Of Modify

Slope Formula: y₂ – y₁ 5 – 2 three
= =
10₂ – ten₁ 3 – ane 2

I promise that this helps you!

Btw, ₁ and ₂ are subscripts, they do not have annihilation to do with the bodily working of the trouble.

The correct answer is 3/2.

Slope = Rate Of Change

Slope Formula: y₂ – y₁ 5 – 2 3
= =
x₂ – x₁ iii – i 2

I hope that this helps you!

Btw, ₁ and ₂ are subscripts, they practise not have anything to do with the actual working of the trouble.

(1,two)(3,5)
slope (rate of change) = (5 – 2) / (3 – 1) = 3/2 <==

$frac{3}{2}$

Step-by-step explanation:

The boilerplate charge per unit of change of a linear function (the gradient) is constant.

The rate of alter of the linear relationship= $frac{text{change in y}}{text{change in x}}$

The rate of alter from x=1 to 10=seven

= $frac{11-2}{7-1}=frac{9}{6}=frac{3}{2}$

Similarly you tin check the rate of change for every interval, it would be

$frac{3}{2}$ .

three over ii

Footstep-past-stride explanation:

We know that the rate of change of a function y=f(x) is given by :-

$k=dfrac{text{change in y}}{text{change in x}}$

From the consecutive values in the tabular array, the change in x = 3-1=2

Change in y = 5-2=3

Now, the rate of modify of the linear relationship modeled in the table will be ;-

$k=dfrac{3}{text{2}}$

~Hello there!

My respond for this question would exist 3/2 because when y'all are determining the rate of change the y ever goes over the 10 value and then in this example apparently 3-1 equals ii and five-two equals 3 and so merely put those values in fraction form and you get 3/ii!!

~Hope this helped have a nice mean solar day!~

As 10 increases by two, y increases by 3, so the rate = modify in y/alter in x = 3/2 = i.5

I retrieve your answer would exist 3/2

Take 2 of ur points…any 2 of them, and so use the gradient formula.

gradient(rate of change) = (y2 – y1) / (x2 – x1)
(1,2)…x1 = 1 and y1 = 2
(3,5)…x2 = 3 and y2 = v
now sub into the formula
gradient(charge per unit of change) = (v – ii) / (3 – 1) = three/2 <==

The answer is D. Three over two