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blagie [28]
3 years ago
9

If you are smart in ranges, graphs, and functions, pls help me out. Look at the screenshot below!

Mathematics
1 answer:
jeka943 years ago
6 0

Answer:

<u>Graph of the equation y = 2/5x - 1</u>

Please check the attached graph of the equation where:

The point  (0, -1) represents the y-intercept, it is the point where the line crosses the y-axis.

The point  (2.5, 0) represents the x-intercept, it is the point where the line crosses the x-axis.

Step-by-step explanation:

Given the equation

y = 2/5x - 1

<u>Determining the y-intercept:</u>

We know that the value of the y-intercept can be determined by setting x = 0 and determining the corresponding value of y.

As the equation is given such as

y = 2/5x - 1

substitute x = 0

y = 2/5(0) - 1

y = 0 - 1

y = -1

so the ordered pair (0, -1) represents the y-intercept.

Thus, the y-intercept is: -1

<u>Determining the x-intercept:</u>

We know that the value of the x-intercept can be determined by setting x = y and determining the corresponding value of x.

As the equation is given such as

y = 2/5x - 1

substitute y = 0

0 = 2/5x - 1

2/5x = 1

x = 5/2

x = 2.5

so the ordered pair (2.5, 0) represents the x-intercept.

<u>Graph of the equation y = 2/5x - 1</u>

Please check the attached graph of the equation where:

The point  (0, -1) represents the y-intercept, it is the point where the line crosses the y-axis.

The point  (2.5, 0) represents the x-intercept, it is the point where the line crosses the x-axis.

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Step-by-step explanation:

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3 years ago
The position of an object along a vertical line is given by s(t) = −t3 + 3t2 + 7t + 4, where s is measured in feet and t is meas
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Answer:

The maximum velocity of the object in the time interval [0, 4] is 10 ft/s.

Step-by-step explanation:

Given : The position of an object along a vertical line is given by s(t) = -t^3+3t^2+7t +4, where s is measured in feet and t is measured in seconds.

To find : What is the maximum velocity of the object in the time interval [0, 4]?

Solution :

The velocity is rate of change of distance w.r.t time.

Distance in terms of t is given by,

s(t) = -t^3+3t^2+7t +4

Derivate w.r.t. time,

v(t)=s'(t) = -3t^2+6t+7

It is a quadratic function so its maximum is at vertex of the function.

The x point of the function is given by,

x=-\frac{b}{2a}

Where, a=-3, b=6 and c=7

t=-\frac{6}{2(-3)}

t=-\frac{6}{-6}

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As 1 lie between interval [0,4]

Substitute t=1 in the function,

v(t)= -3(1)^2+6(1)+7

v(t)= -3+6+7

v(1)=10

Th maximum velocity is 10 ft/s.

Therefore, the maximum velocity of the object in the time interval [0, 4] is 10 ft/s.

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3 years ago
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