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Tomtit [17]
3 years ago
11

Sherese draws a segment and labels the endpoints R and S. She uses a compass and straightedge to construct a segment congruent t

o RS¯¯¯¯¯ .
Drag and drop the steps Sherese uses in order from start to finish.

Mathematics
2 answers:
Lisa [10]3 years ago
8 0

Answer:

1. Adjust the compass opening to the width of RS

2. Draw point P anywhere

3. Place the compass point on point P and draw an arc.

4. Label any point on the arc as point Q

5. Use a straightedge to connect points P and Q

Step-by-step explanation:

I did the test. This was the right answer.

dangina [55]3 years ago
5 0

Answer:

Adjust the cpmpass opening to the width of RS

Draw point P anywhere

Place the compass point on point p and draw an arc

label any points on the arc as point Q

use a straightedge to connect points p and Q

Step-by-step explanation:

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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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