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Colt1911 [192]
2 years ago
14

Jorden drives to the store at 30 miles per hour. On her way home she averages only 20 miles per hour. If the total driving time

takes half an hour, how far does she live from the store?
Mathematics
2 answers:
kozerog [31]2 years ago
7 0

Answer:

Rate column: 30, 20.

Distance column: x miles for both.

Step-by-step explanation:

Trava [24]2 years ago
4 0
For this case we have the following definition:
 d = v * t

 Where,
 d: distance
 v: speed
 t: time
 Clearing the time we have:
 t = d / v

 The total driving time takes half an hour:
 t = d / v1 + d / v2

1.5 = d / 30 + d / 20
 Clearing d we have:
 1.5 * 30 * 20 = 20d + 30d

1.5 * 30 * 20 = 50d

d = (1.5 * 30 * 20) / (50)

d = 18 miles
 Answer:
 
she lives 18 miles far from the store
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Answer:

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Step-by-step explanation:

given data

volume = 150 in³

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remainder sides cost = 2 cents/in²

to find out

the dimensions that will minimize the cost

solution

we consider here length = x and breadth = y and height = z

and

area of base = xy

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and area of remaining side = xz + 2yz     .....................1

so

cost of base will be = 5xy

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cost of remaining side = 2 ( xz+ 2yz)        

and

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total cost TC = 5xy + 10xz + 2 ( xz+ 2yz)  

total cost TC = 5xy + 10xz + 2xz+ 4yz

total cost TC = 5xy + 12xz + 4yz                  ....................2

and total volume will be = xyz

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now put z value in equation 2

total cost TC = 5xy + 12xz + 4yz    

total cost TC = 5xy + 12x\frac{150}{xy} + 4y\frac{150}{xy}

total cost TC = 5xy + \frac{1800}{y} + \frac{600}{x}     ...........4

now differentiate TC w.r.t x and y

TC (x) = 5y - \frac{600}{x^2}

TC (x) = 5x - \frac{1800}{y^2}

now equating with 0 these

5y - \frac{600}{x^2} = 0

x² = \frac{120}{y}    

and

5x - \frac{1800}{y^2} = 0

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z = \frac{150}{xy}  

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so minimum cost of construction box are x = x = 3.4199 in and y = 10.2598 in and z = 4.2750 in

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