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NeX [460]
4 years ago
11

•Pls help

Mathematics
1 answer:
kramer4 years ago
7 0

Answer:

The recipe makes 1/2 cup of salad dressing assuming that the salt and herbs do not make a difference.  Each serving has 1/12 of a cup.  I am not gifted in the number sentence and equation department

Step-by-step explanation:

1/3=2/6

         +

1/6=1/6

--------------

3/6=1/2

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The figure is made up of a cylinder and a hemisphere. To the nearest whole number, what is the approximate volume of this figure
Novosadov [1.4K]

Hello!

The figure is made up of a cylinder 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.  in.³

Data: (Cylinder)

h (height) = 7 in

r (radius) = 2.5 in (The diameter is 5 being twice the radius)

Adopting: \pi \approx 3.14

V (volume) = ?

Solving: (Cylinder volume)

V = \pi *r^2*h

V = 3.14 *2.5^2*7

V = 3.14*6.25*7

V = 137.375 \to \boxed{V_{cylinder} \approx 137.38\:in^3}

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

Data: (hemisphere volume)

V (volume) = ?

r (radius) = 2.5 in (The diameter is 5 being twice the radius)

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{2.5^3}{3}

V = 2*3.14 * \dfrac{15.625}{3}

V = \dfrac{98.125}{3}

\boxed{ V_{hemisphere} \approx 32.70\:in^3}

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

Volume of the figure = cylinder volume + hemisphere volume

Volume of the figure = 137.38 in³ + 32.70 in³

Volume\:of\:the\:figure =170.08 \to \boxed{\boxed{\boxed{Volume\:of\:the\:figure = 170\:in^3}}}\end{array}}\qquad\quad\checkmark

_______________________

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

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