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olga2289 [7]
3 years ago
8

Radius of cylinder = 5 cm

Mathematics
2 answers:
Kisachek [45]3 years ago
7 0

Answer:

C  1055 04 cm

Step-by-step explanation:

We don't need to see the figure, since we know for sure the cone fits into the cylinder (smaller diameter and height).

So, we first need to calculate the volume of the cylinder, which is given by the formula:

VT = π * r² * h

VT = 3.14 * 5² * 16 = 3.14 * 400 = 1,256 cubic cm

Then we calculate the volume of the cone, which is given by:

VC = (π * r² * h)/3

VC = (3.14 * 4² * 12)/3 = (3.14 * 192)/3 = 200.96 cu cm

Then we calculate the void space left inside the cylinder by subtracting the volume of the cone from the volume of the cylinder:

NV = VT - VC = 1,256 - 200.96 = 1,055.04 cu cm

kramer3 years ago
4 0

Answer:

C)1055.04cm^3

Step-by-step explanation:

(picture explains)

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2x+3y=1 and y=-2 - 9 solve using linear combination method. Please explain steps.
kaheart [24]

Answer:

x=17

Step-by-step explanation:

2x+3y=1

y=-2-9

  • In order to combine these two equations, an idea you need to keep in mind is finding a way of setting these equations as equal to each other. I saw that each equation shared a common value, y. In this case, we need to isolate y in the first equation so that both equations =y.

2x+3y=1

3y=-2x+1

\frac{3y}{3}=\frac{-2x+1}{3}

y=-\frac{2}{3}x+\frac{1}{3}

  • With this, we now know that both -2-9 and -\frac{2}{3}x+\frac{1}{3} are equal to y, so we can set them equal to each other.

y=-\frac{2}{3}x+\frac{1}{3}

y=-11

-\frac{2}{3}x+\frac{1}{3}=-11

  • Reply to this if anything I'm saying or doing is confusing in any way, or if you find a mistake. :) Solve for x.

-\frac{2}{3}x+\frac{1}{3}=-11

-\frac{2}{3}x=-11-\frac{1}{3}

-\frac{2}{3}x =-\frac{33}{3}-\frac{1}{3}

-\frac{2}{3}x =-\frac{34}{3}

-\frac{2}{3}x(-\frac{3}{2})=-\frac{34}{3}(-\frac{3}{2})

x=\frac{102}{6}=17

x=17

  • Hopefully this answer is correct AND makes sense in terms of how I achieved it. Again, reply to this with any questions or mistakes I made and I'll do my best to answer or fix them.

 

4 0
3 years ago
Please help i don’t understand this
sergij07 [2.7K]

yes it does have enough to cover the cube

8 0
3 years ago
What does P(X4=2,X3=3,X2=3,X1=2)
Dmitriy789 [7]

Answer:

its b

Step-by-step explanation:

3 0
3 years ago
Prove that the segments joining the midpoint of consecutive sides of an isosceles trapezoid form a rhombus.
sergiy2304 [10]

Answer:

See explanation

Step-by-step explanation:

a) To prove that DEFG is a rhombus, it is sufficient to prove that:

  1. All the sides of the rhombus are congruent:  |DG|\cong |GF| \cong |EF| \cong |DE|
  2. The diagonals are perpendicular

Using the distance formula; d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

|DG|=\sqrt{(0-(-a-b))^2+(0-c)^2}

\implies |DG|=\sqrt{a^2+b^2+c^2+2ab}

|GF|=\sqrt{((a+b)-0)^2+(c-0)^2}

\implies |GF|=\sqrt{a^2+b^2+c^2+2ab}

|EF|=\sqrt{((a+b)-0)^2+(c-2c)^2}

\implies |EF|=\sqrt{a^2+b^2+c^2+2ab}

|DE|=\sqrt{(0-(-a-b))^2+(2c-c)^2}

\implies |DE|=\sqrt{a^2+b^2+c^2+2ab}

Using the slope formula; m=\frac{y_2-y_1}{x_2-x_1}

The slope of EG is m_{EG}=\frac{2c-0}{0-0}

\implies m_{EG}=\frac{2c}{0}

The slope of EG is undefined hence it is a vertical line.

The slope of  DF is m_{DF}=\frac{c-c}{a+b-(-a-b)}

\implies m_{DF}=\frac{0}{2a+2b)}=0

The slope of DF is zero, hence it is a horizontal line.

A horizontal line meets a vertical line at 90 degrees.

Conclusion:

Since |DG|\cong |GF| \cong |EF| \cong |DE| and DF \perp FG , DEFG is a rhombus

b) Using the slope formula:

The slope of DE is m_{DE}=\frac{2c-c}{0-(-a-b)}

m_{DE}=\frac{c}{a+b)}

The slope of FG is m_{FG}=\frac{c-0}{a+b-0}

\implies m_{FG}=\frac{c}{a+b}

5 0
3 years ago
The ratio of boys to girls in weekend volunteers is 5 to 6. There are 24 girls in the group. How many boys are there?
Zanzabum

there are 20 boys in the group.


5 0
3 years ago
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