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True [87]
3 years ago
11

Nour drove from the Dead Sea up to Amman and her altitude was 409 meters below sea level. When she arrived in Amman 2 hours late

r, her altitude was 1000 meters above sea level. Let y represent Nour's altitude(in meters) relative to sea level after x hours
Mathematics
1 answer:
kozerog [31]3 years ago
6 0

Answer:

y=  704.5x - 409

Step-by-step explanation:

Complete question is : . Complete the equation for the relationship between the altitude and number of hours.

SOLUTION

By plotting altitude on y-axis and time or duration of travel on x-axis of the graph, we will have the the location of Nour at t = 0 will be represented by (0, -409)

when t = 2 hours location points will be  at (2, 1000).

As you know, at a constant rate her altitude is increasing , we will have a linear graph with equation y = mx + b --> eq(1)

Slope of the line is given by:

m= (y-y') / (x-x') => (1000+409)/(2-0)

m= 704.5

y-intercept 'b' is -409

Therefore, the eq(1) will become

(1)=> y=  704.5x - 409

Where, 'y' represents Nour's altitude(in meters) relative to sea level after 'x' hours

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3 years ago
The graph h = −16t^2 + 25t + 5 models the height and time of a ball that was thrown off of a building where h is the height in f
Thepotemich [5.8K]

Answer:

part 1) 0.78 seconds

part 2) 1.74 seconds

Step-by-step explanation:

step 1

At about what time did the ball reach the maximum?

Let

h ----> the height of a ball in feet

t ---> the time in seconds

we have

h(t)=-16t^{2}+25t+5

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

The vertex represent a maximum

so

The x-coordinate of the vertex represent the time when the ball reach the maximum

Find the vertex

Convert the equation in vertex form

Factor -16

h(t)=-16(t^{2}-\frac{25}{16}t)+5

Complete the square

h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+5+\frac{625}{64}

h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}\\h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}

Rewrite as perfect squares

h(t)=-16(t-\frac{25}{32})^{2}+\frac{945}{64}

The vertex is the point (\frac{25}{32},\frac{945}{64})

therefore

The time when the ball reach the maximum is 25/32 sec or 0.78 sec

step 2

At about what time did the ball reach the minimum?

we know that

The ball reach the minimum when the the ball reach the ground (h=0)

For h=0

0=-16(t-\frac{25}{32})^{2}+\frac{945}{64}

16(t-\frac{25}{32})^{2}=\frac{945}{64}

(t-\frac{25}{32})^{2}=\frac{945}{1,024}

square root both sides

(t-\frac{25}{32})=\pm\frac{\sqrt{945}}{32}

t=\pm\frac{\sqrt{945}}{32}+\frac{25}{32}

the positive value is

t=\frac{\sqrt{945}}{32}+\frac{25}{32}=1.74\ sec

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A rectangle has an area of 100 square centimeters. The base of the rectangle is 25 centimeters. What is the height of the rectan
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