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3 years ago

Explain how to determine if a proportional relationship exists between two variables in a graph

1 answer:

Natasha2012 [34]3 years ago

8 0

Here are some steps to determine if a proportional relationship exists between two variables in a graph ! Hope it's useful~

*1. Note whether the line is straight. When two variables are in proportion, the line representing them will be straight. This means that the slope of the line is constant, or follows the equation\

*2. Determine the y-intercept. The y-intercept is the point where the line crosses the y-axis. When two variables are directly proportional, when graphed their line will cross through the origin. The origin is at the point {(0,0)}, so the y-intercept of the line should be (0). If it isn’t, the variables are not directly proportional.**The y-axis is the vertical axis.

*3. Find the coordinates of two points on the line. Compare the coordinates with each other, and determine whether each coordinate changed by the same factor.[6] That is, determine whether the constant ({k}) is the same for both the { x} and { y} values.For example, if the first point is {(1,3)}, and the second point is { (2,6)}, the x-coordinate changed by a factor of 2, since { 1(2)=2}. The y-coordinate also changed by a factor of 2, since { 3(2)=6}. Thus, you can confirm that the line represents two variables that are directly proportional.

Best Regards,,,
Wish it was the answer you're looking for !!
Information taken from the following web:
https://www.wikihow.com/Determine-Whether-Two-Variables-Are-Directly-Proportional#Using_a_Graph_sub

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3 years ago

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OlgaM077 [116]

Answer: I believe that the answer is B

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3 years ago

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SOLVE THE QUESTION BELOW ASAP

qwelly [4]

Answer:

Part A) The graph in the attached figure (see the explanation)

Part B) 16 feet

Part C) see the explanation

Step-by-step explanation:

Part A) Graph the function

Let

h(t) ----> the height in feet of the ball above the ground

t -----> the time in seconds

we have

$h(t)=-16t^{2}+98$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex is a maximum

To graph the parabola, find the vertex, the intercepts, and the axis of symmetry

Find the vertex

The function is written in vertex form

The vertex is the point (0,98)

Find the y-intercept

The y-intercept is the value of the function when the value of t is equal to zero

For t=0

$h(t)=-16(0)^{2}+98$

$h(0)=98$

The y-intercept is the point (0,98)

Find the t-intercepts

The t-intercepts are the values of t when the value of the function is equal to zero

For h(t)=0

$-16t^{2}+98=0$

$t^{2}=\frac{98}{16}$

square root both sides

$t=\pm\frac{\sqrt{98}}{4}$

$t=\pm7\frac{\sqrt{2}}{4}$

therefore

The t-intercepts are

$(-7\frac{\sqrt{2}}{4},0), (7\frac{\sqrt{2}}{4},0)$

$(-2.475,0), (2.475,0)$

Find the axis of symmetry

The equation of the axis of symmetry in a vertical parabola is equal to the x-coordinate of the vertex

$x=0$ ----> the y-axis

To graph the parabola, plot the given points and connect them

we have

The vertex is the point $(0,98)$

The y-intercept is the point $(0,98)$

The t-intercepts are $(-2.475,0), (2.475,0)$

The axis of symmetry is the y-axis

The graph in the attached figure

Part B) How far is the artifact fallen from the time t=0 to time t=1

we know that

For t=0

$h(t)=-16(0)^{2}+98$

$h(0)=98\ ft$

For t=1

$h(t)=-16(1)^{2}+98$

$h(1)=82\ ft$

Find the difference

$98\ ft-82\ ft=16\ ft$

Part C) Does the artifact fall the same distance from time t=1 to time t=2 as it does from the time t=0 to time t=1?

we know that

For t=1

$h(t)=-16(1)^{2}+98$

$h(1)=82\ ft$

For t=2

$h(t)=-16(2)^{2}+98$

$h(2)=34\ ft$

Find the difference

$82\ ft-34\ ft=48\ ft$

The artifact fall 48 feet from time t=1 to time t=2 and fall 16 feet from time t=0 to time t=1

therefore

The distance traveled from t=1 to t=2 is greater than the distance traveled from t=0 to t=1

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Paraphin [41]

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zhenek [66]

Answer:

$\rule6cm99999cm$

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2 years ago

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