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3 years ago

PLEASE HELP!!!!

Mathematics

1 answer:

fenix001 [56]3 years ago

4 0

Answer:

parallelogram

Step-by-step explanation:

Looks like a rectangle, but then angles should be marked =90

But two sides are marked equal that makes it a parallelogram.

(a parallelogram is 4 sided figure where 2 opposite sides are =

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PLS HELP ME!!! The figure is made up of a cone and a hemisphere. To the nearest whole number, what is the approximate volume of

Pachacha [2.7K]

Hello!

The figure is made up of a cone and a hemisphere. To the nearest whole number, what is the approximate volume of this figure? Use 3.14 to approximate π . Enter your answer in the box. cm³

Data: (Cone)

h (height) = 12 cm

r (radius) = 4 cm (The diameter is 8 being twice the radius)

Adopting: $\pi \approx 3.14$

V (volume) = ?

Solving: (Cone volume)

$V = \dfrac{ \pi *r^2*h}{3}$

$V = \dfrac{ 3.14 *4^2*\diagup\!\!\!\!\!12^4}{\diagup\!\!\!\!3}$

$V = 3.14*16*4$

$\boxed{V = 200.96\:cm^3}$

Note: Now, let's find the volume of a hemisphere.

Data: (hemisphere volume)

V (volume) = ?

r (radius) = 4 cm

Adopting: $\pi \approx 3.14$

If: We know that the volume of a sphere is $V = 4* \pi * \dfrac{r^3}{3}$ , but we have a hemisphere, so the formula will be half the volume of the hemisphere $V = \dfrac{1}{2}* 4* \pi * \dfrac{r^3}{3} \to \boxed{V = 2* \pi * \dfrac{r^3}{3}}$

Formula: (Volume of the hemisphere)

$V = 2* \pi * \dfrac{r^3}{3}$

Solving:

$V = 2* \pi * \dfrac{r^3}{3}$

$V = 2*3.14 * \dfrac{4^3}{3}$

$V = 2*3.14 * \dfrac{64}{3}$

$V = \dfrac{401.92}{3}$

$\boxed{ V_{hemisphere} \approx 133.97\:cm^3}$

What is the approximate volume of this figure?

Now, to find the total volume of the figure, add the values: (cone volume + hemisphere volume)

Volume of the figure = cone volume + hemisphere volume

Volume of the figure = 200.96 cm³ + 133.97 cm³

$\boxed{\boxed{\boxed{V = 334.93\:cm^3 \to Volume\:of\:the\:figure \approx 335\:cm^3 }}}\end{array}}\qquad\quad\checkmark$

Answer:

The volume of the figure is approximately 335 cm³

_______________________

I Hope this helps, greetings ... Dexteright02! =)

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