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eduard
3 years ago
15

A basketball court is rectangular and measures 28 meters long and 15meters wide. How long is the diagonal of a basketball court

(to the nearest 10th)
Mathematics
1 answer:
FrozenT [24]3 years ago
8 0
All you do is find the are which would be a=l×w or a=28×15 or 420 or to the nearest 10th would be also 420
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If it is 4:40 what time would it be a half of hour later
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6:10.

Step-by-step explanation:

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3 years ago
"A rectangle has a height of 5 cm and its base is increasing at a rate" of 3/2 cm/min. When its area is 60 cm2, at what rate is
sukhopar [10]

Answer:

The diagonal is increasing at the rate of 119/104cm/min of the given rectangle.

Step-by-step explanation:

Dimensions of the rectangle

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Area = 60cm^2

We know the area of a rectangle of given by = base* Height

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b = 12cm

Applying Pythagoras theorem while drawing a diagonal to the rectangle

  b^2 +h^2 =  D^2\\

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so our diagonal will be 13cm  

Upon differentiating the area of the rectangle  we get

b*h = A=60cm^2

using  the chain rule of differentiation

h*db/dt + b*dh/dt  = 0

b*dh/dt = -h*db/dt

12*dh/dt = -5*3/2

dh/dt = -5/8 cm//min

so the height of the rectangle is decreasing at the rate of -5/8cm/min

now we have all the measurements we need

b = 12 , db/dt = 3/2cm/min

h = 5 , dh/dt = -5/8 cm/min

b^2 +h^2  = D^2

Upon differentiating we get

2b*db/dt + 2h*dh/dt = 2D*dD/dt

b*db/dt + h*dh/dt = D*dD/dt

12*3/2 + 5*(-5/8) = 13*dD/dt

18 -25/8 = 13*dD/dt

\frac{144-25}{8} = 13*dD/dt

dD/dt = \frac{119}{104} cm/min

Therefore the diagonal is increasing at the rate of 119/104cm/min of the given rectangle.

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