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Basile [38]
3 years ago
4

A box is a cuboid with dimensions 28cm by 15cm by 20cm all measured to the nearest centimetre.

Mathematics
1 answer:
mariarad [96]3 years ago
7 0

Answer:

The 17 disc cases would definitely fit into the box.

Step-by-step explanation:

The given cuboid box has the dimensions 28cm by 15cm by 20cm.

Disc cases are cuboid with dimensions 1.5cm by 14.2cm by 19.3cm.

volume of a cuboid = length × width × height

Volume of the box = 28 × 15 × 20

                               = 8400 cubic centimeters

Volume of each disc case = 1.5 × 14.2 × 19.3

                                           = 411.09 cubic centimeters

When the 17 disc cases are stacked it would have a volume.

The volume of 17 disc cases = 17 × volume of a case

                                               = 17 × 411.09

                                               = 6988.53 cubic centimeters

Thus comparing the volume for 17 disc cases and that of the cuboid box, the disc cases would definitely fit into the box.

i.e           =   \frac{volume of box}{volume of 17 disc cases}  

              = \frac{8400}{6988.53}

             = 1.20

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A man starts walking north at 2 ft/s from a point P. Five minutes later a woman starts walking south at 3 ft/s from a point 500
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Answer:

The rate at which both of them are moving apart is 4.9761 ft/sec.

Step-by-step explanation:

Given:

Rate at which the woman is walking,\frac{d(w)}{dt} = 3 ft/sec

Rate at which the man is walking,\frac{d(m)}{dt} = 2 ft/sec

Collective rate of both, \frac{d(m+w)}{dt} = 5 ft/sec

Woman starts walking after 5 mins so we have to consider total time traveled by man as (5+15) min  = 20 min

Now,

Distance traveled by man and woman are m and w ft respectively.

⇒ m=2\ ft/sec=2\times \frac{60}{min} \times 20\ min =2400\ ft

⇒ w=3\ ft/sec = 3\times \frac{60}{min} \times 15\ min =2700\  ft

As we see in the diagram (attachment) that it forms a right angled triangle and we have to calculate \frac{dh}{dt} .

Lets calculate h.

Applying Pythagoras formula.

⇒ h^2=(m+w)^2+500^2  

⇒ h=\sqrt{(2400+2700)^2+500^2} = 5124.45

Now differentiating the Pythagoras formula we can calculate the rate at which both of them are moving apart.

Differentiating with respect to time.

⇒ h^2=(m+w)^2+500^2

⇒ 2h\frac{d(h)}{dt}=2(m+w)\frac{d(m+w)}{dt}  + \frac{d(500)}{dt}

⇒ \frac{d(h)}{dt} =\frac{2(m+w)\frac{d(m+w)}{dt} }{2h}                         ...as \frac{d(500)}{dt}= 0

⇒ Plugging the values.

⇒ \frac{d(h)}{dt} =\frac{2(2400+2700)(5)}{2\times 5124.45}                       ...as \frac{d(m+w)}{dt} = 5 ft/sec

⇒ \frac{d(h)}{dt} =4.9761  ft/sec

So the rate from which man and woman moving apart is 4.9761 ft/sec.

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