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Ray Of Light [21]
3 years ago
13

9. What formula could be used to find the area of the image?

Mathematics
1 answer:
Scilla [17]3 years ago
8 0

Answer:

9. Area of triangle= \frac{1}{2} \times base\times height

10. Base= 16 ft and Height= 15 ft.

11. 120 ft²

Step-by-step explanation:

Given: Base of triangle= 16 ft

           Length of leg of triangle= 17 ft.

As shown in the picture that median has bisected the triangle and forming two right angle triangle.

∴ Each base of right angle triangle= \frac{16}{2} = 8\ feet

Now, finding the height of triangle by using pythagorean theorem.

We know, h^{2} = a^{2} +b^{2}, where h is hypotenous, a is height or opposite and b is base or adjacent of triangle.

⇒ 17^{2}= a^{2} + 8^{2}

⇒289= a^{2} + 64

Subtracting both side by 64

⇒ 225= a^{2}

taking square root on both side, remember; √a²=a

⇒a= \sqrt{225} = 15

Hence, height of triangle is 15 ft.

Next, finding the area of triangle.

Formula; Area of triangle= \frac{1}{2} \times base\times height

⇒ Area of triangle= \frac{1}{2} \times 16\times 15

∴ Area of triangle= 120 ft²

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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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