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denis23 [38]
3 years ago
13

Randy walks 1/2 mile in each 1/5 hour. How far will Randy walk in one hour

Mathematics
1 answer:
katrin [286]3 years ago
7 0
In the given problem, it is already given that Randy can walk for a distance of 1/2 a mile in 1/5 of an hour. Now let us first convert 1/5 hour to minutes. this will make the problem a bit easy.
1/5 hour = 1/5 * 60 minutes
               = 12 minutes
So
Randy in 12 minutes can walk for a distance of = 1/2 mile
Then
In 60 minutes Randy can walk for a distance of = [1/(2*12)] * 60
                                                                             = (1/24) * 60
                                                                              = 2.5
So Randy can walk 2.5 miles in 1 hour.
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here a subtraction problem that has been solved incorrectly what is the right answer for 74-25=59 but do it in the addition sent
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Answer:

the equation is wrong

Step-by-step explanation:

=> 74-25=59

since 74 - 25 equals to 59 in order to find the addition sentence that will check the subtraction problem is true and 59 and 3/5 or you can add add 59 and 25 to check if the problem was solved correctly

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3 years ago
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A ​20-ft by ​the 40-ft rectangular swimming pool is surrounded by a walkway of uniform width. If the total area of the walkway i
Jlenok [28]
<h3>Answer:  2 feet</h3>

========================================================

Explanation:

x = width of the walkway in feet

This is some positive real number.

The dimension of 20 feet bumps up to 20+2x when adding on x from both directions. Similarly, the 40 ft dimension becomes 40+2x

Refer to the diagram below.

The 20 ft by 40 ft pool is surrounded by a larger rectangle that is 20+2x ft by 40+2x ft

The pool itself is 20*40 = 800 sq ft. Add on the walkway area to get 800+256 = 1056 sq ft.

-----------

Area = length*width

1056 = (20+2x)*(40+2x)

1056 = 20*40 + 20*2x + 2x*40 + 2x*2x ... FOIL rule

1056 = 800 + 40x + 80x + 4x^2

0 = 4x^2 + 40x + 80x + 800 - 1056

0 = 4x^2 + 120x - 256

4x^2 + 120x -  256 = 0

4(x^2 + 30x - 64) = 0

x^2 + 30x - 64 = 0

Let's use the quadratic formula to finish solving for x.

Plug in a = 1, b = 30, c = -64

x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-30\pm\sqrt{(30)^2-4(1)(-64)}}{2(1)}\\\\x = \frac{-30\pm\sqrt{1156}}{2}\\\\x = \frac{-30\pm34}{2}\\\\x = \frac{-30+34}{2} \ \text{ or } \ x = \frac{-30-34}{2}\\\\x = \frac{4}{2} \ \text{ or } \ x = \frac{-64}{2}\\\\x = 2 \ \text{ or } \ x = -32\\\\

Recall we made x be positive. This is because a negative walkway width does not make sense. This means we'll ignore x = -32.

The only practical solution is x = 2

Therefore, the walkway is <u>2 feet wide</u>

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