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statuscvo [17]
3 years ago
12

Mito bakes biscuits.He uses 1/4 cup of flour to coat the countertop and the rolling pin.He also uses 2 1/2 cups of flour for eac

h batch of biscuits he bakes.If he uses 7 3/4 cups of flour in all,how many batches of biscuits does he bake.Write and solve an equation
Mathematics
1 answer:
Hoochie [10]3 years ago
5 0

Answer:

<u>3  batches of biscuits.</u>

Step-by-step explanation:

For one batch of biscuits, Mito uses 1/4 cup of flour to coat the countertop and the rolling pin.He also uses 2 1/2 cups of flour for each batch of biscuits he bakes

When he uses 7 3/4 cups of flour

Let the number of batches = x

convert the given data to decimal:

2 1/2 = 2.5   and   1/4 = 0.25    and  7 3/4 = 7.75

So, the equation representing the problem is:

2.5 x + 0.25 = 7.75

Solve for x

2.5x = 7.75 - 0.25 = 7.5

x=7.5/2.5 = 75/25 = 3

<u>So, the number of batches of biscuits = 3 batches.</u>

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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}

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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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