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ZanzabumX [31]
3 years ago
13

A won a race at the local fair by running 10.5 miles in exactly 2 hours. At this constant​ rate, how long does it take the same

dog to run the 21 mile state fair​ race? Use ratio reasoning to solve. * 1 point
Mathematics
1 answer:
Luba_88 [7]3 years ago
3 0

Answer:

The dog will run the 21-mile race in 4.04 hours

Step-by-step explanation:

The first thing we need to do is find out how far he ran the race at the local fair in 1 hour.

To do this, we can set up a simple relationship:

If he ran 10.5 miles in 2 hours,

He will run x miles in 1 hour

Cross multiplying, we have

x = (10.5 X 1)/2 = 5.25 miles in 1 hour.

We can now comfortably say that he runs at the rate of 5.2miles/hr

He now goes to the state fair race, which is 21 miles long.

if he runs at the rate of 5.2 miles in 1 hour,

He will run 21 miles in y hours

y = 21 / 5.2 = 4.04 hours

The dog will run the 21-mile race in 4.04 hours

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1/(8^1) or 1/8 is your answer

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The length of a rectangle is 7 inches more than its width. The area of the rectangle is equal to 2 inches less than 5 times the
fredd [130]
Let the width be w, then the length is w+7 units.

The area of the rectangle is A=w(w+7)= w^{2}+7w.

The perimeter of the rectangle is: 

P = 2(Width + Length)=2(w+w+7)=2(2w+7)=4w+14

"The area of the rectangle is equal to 2 inches less than 5 times the perimeter." means that:

A = 5P - 2

w^{2}+7w=5(4w+14)-2

w^{2}+7w=20w+70-2

w^{2}-13w-68=0

to solve the quadratic equation, let's use the quadratic formula

let a=1, b=-13, c=-68

D= b^{2} -4ac=169-4(1)(-68)=169+272=441

the root of the discriminant is 21

the roots are

w1=(13+21)/2=34/2=17
and
w2=(13-21)/2=-8/2=-4, which cannot be the width.

The width is 17 units, and the length is 17+7=24 units


Answer: w=17, l=24
6 0
2 years ago
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