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ANEK [815]
3 years ago
8

George walks 1 mile to school. He leaves home at the same time each day, walks at a steady speed of 3 miles per hour, and arrive

s just as school begins. Today he was distracted by the pleasant weather and walked the first 1/2 mile at a speed of only 2 miles per hour. At how many miles per hour must George run the last 1/2 mile in order to arrive just as school begins today?
Mathematics
1 answer:
alex41 [277]3 years ago
4 0

Answer:

George must run the last half mile at a speed of 6 miles per hour in order to arrive at school just as school begins today

Step-by-step explanation:

Here, we are interested in calculating the number of hours George must walk to arrive at school the normal time he arrives given that his speed is different from what it used to be.

Let’s first start at looking at how many hours he take per day on a normal day, all things being equal.

Mathematically;

time = distance/speed

He walks 1 mile at 3 miles per hour.

Thus, the total amount of time he spend each normal day would be;

time = 1/3 hour or 20 minutes

Now, let’s look at his split journey today. What we know is that by adding the times taken for each side of the journey, he would arrive at the school the normal time he arrives given that he left home at the time he used to.

Let the unknown speed be x miles/hour

Mathematically;

We shall be using the formula for time by dividing the distance by the speed

1/3 = 1/2/(2) + 1/2/x

1/3 = 1/4 + 1/2x

1/2x = 1/3 - 1/4

1/2x = (4-3)/12

1/2x = 1/12

2x = 12

x = 12/2

x = 6 miles per hour

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A toy rocket is launched vertically from 5 feet above ground level with an initial velocity of 112 feet per second. The height h
Masja [62]

Answer:

Step-by-step explanation:

a)The height h after t seconds is given by the equation h(t)=-16t^2+112t+5.

Where 5 represents 5feet above the ground and this is the height from which the rocket was launched.

The equation is a quadratic equation. The plot of this equation on a graph would give a parabola whose vertex would be equal to the maximum height travelled by the rocket.

The vertex of the parabola is calculated as follows,

Vertex = -b/2a

From the equation,

a = -16

b = 112

Vertex = - - 112/32= 3.5

So the rocket will attain maximum height at 3.5 seconds.

It will also take 3.5 seconds to reach the ground. This means it will take a total of 3.5 seconds + 3.5 seconds

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b) to find time it will take to be 100 feet above the ground,

-16t^2+112t-95 = 0

Using general quadratic equation formula,

a = -16

b = 112

c = -95

t = [-b+/-√b^2-4ac]/2a

t = [ -112 +/-√112^2-4×-16×-95]/16 × -2

= [-112 +/-√12544-6080]/-32

= (-112+80.4)/-32 or (-112-80.4)/-32

t = 0.9875 or t = 6.0125

So it will take 0.9875 or approximately 1 second to be 100 feet above the ground

7 0
3 years ago
P(x)=-4x-2<br> Find p(a-3)
Marta_Voda [28]
Answer : -4a+10

Substitute (a-3) for x
-4(a-3)-2
-4a+12-2
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7 0
3 years ago
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RSB [31]

Answer:

9c² -30c + 25

Step-by-step explanation:

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(3c - 5)²

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Final answer: 9c² -30c + 25

Hope this helps!

7 0
2 years ago
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Answer:

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3 0
2 years ago
Is 0.2487 a interger
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1).  There is no such thing as an ' interger ',
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7 0
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