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yan [13]
2 years ago
10

Jaleel's house is 18 feet high. Jaleel, who is 5 feet tall, stands next to his house on a sunny day. How long is the shadow of J

aleel's house if Jaleel's shadow is 8 feet long?
Mathematics
1 answer:
viktelen [127]2 years ago
4 0

Answer:

21 feet

Step-by-step explanation:

First you find the difference between Jadeel's height and Jadeel's shadow

8-5=3

So the difference is 3

Then you add it to the house shadow because it is how much extra foot high it is

So the house shadow is 21 feet

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Math question
strojnjashka [21]

Answer:

The candle has a radius of 8 centimeters and 16 centimeters and uses an amount of approximately 1206.372 square centimeters.

Step-by-step explanation:

The volume (V), in cubic centimeters, and surface area (A_{s}), in square centimeters, formulas for the candle are described below:

V = \pi\cdot r^{2}\cdot h (1)

A_{s} = 2\pi\cdot r^{2} + 2\pi\cdot r \cdot h (2)

Where:

r - Radius, in centimeters.

h - Height, in centimeters.

By (1) we have an expression of the height in terms of the volume and the radius of the candle:

h = \frac{V}{\pi\cdot r^{2}}

By substitution in (2) we get the following formula:

A_{s} = 2\pi \cdot r^{2} + 2\pi\cdot r\cdot \left(\frac{V}{\pi\cdot r^{2}} \right)

A_{s} = 2\pi \cdot r^{2} +\frac{2\cdot V}{r}

Then, we derive the formulas for the First and Second Derivative Tests:

First Derivative Test

4\pi\cdot r -\frac{2\cdot V}{r^{2}} = 0

4\pi\cdot r^{3} - 2\cdot V = 0

2\pi\cdot r^{3} = V

r = \sqrt[3]{\frac{V}{2\pi} }

There is just one result, since volume is a positive variable.

Second Derivative Test

A_{s}'' = 4\pi + \frac{4\cdot V}{r^{3}}

If \left(r = \sqrt[3]{\frac{V}{2\pi}}\right):

A_{s} = 4\pi + \frac{4\cdot V}{\frac{V}{2\pi} }

A_{s} = 12\pi (which means that the critical value leads to a minimum)

If we know that V = 3217\,cm^{3}, then the dimensions for the minimum amount of plastic are:

r = \sqrt[3]{\frac{V}{2\pi} }

r = \sqrt[3]{\frac{3217\,cm^{3}}{2\pi}}

r = 8\,cm

h = \frac{V}{\pi\cdot r^{2}}

h = \frac{3217\,cm^{3}}{\pi\cdot (8\,cm)^{2}}

h = 16\,cm

And the amount of plastic needed to cover the outside of the candle for packaging is:

A_{s} = 2\pi\cdot r^{2} + 2\pi\cdot r \cdot h

A_{s} = 2\pi\cdot (8\,cm)^{2} + 2\pi\cdot (8\,cm)\cdot (16\,cm)

A_{s} \approx 1206.372\,cm^{2}

The candle has a radius of 8 centimeters and 16 centimeters and uses an amount of approximately 1206.372 square centimeters.

3 0
3 years ago
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