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3 years ago

Everytime Evan walks his neighbors. He gets $8 and he has four or five times and there's a video game

Mathematics

1 answer:

Mariulka [41]3 years ago

8 0

So what’s the question

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from right circular cylinder with height 10 cm and radius of base 6 CM a right circular cone of the same height and base is remo

djyliett [7]

Answer:

753.98 cubic cm.

Step-by-step explanation:

We have been given that from right circular cylinder with height 10 cm and radius of base 6 cm, a right circular cone of the same height and base is removed.

To find the area of remaining solid we will subtract volume of cone from volume of cylinder.

$\text{Volume of remaining solid}=\text{Volume of cylinder- Volume of cone}$

$\text{Volume of remaining solid}=\pi r^{2} h- \frac{1}{3}\pi r^{2}h$

$\text{Volume of remaining solid}=\frac{3}{3}\pi r^{2} h- \frac{1}{3}\pi r^{2}h$

$\text{Volume of remaining solid}=\frac{2}{3}\pi r^{2} h$

Upon substituting our given values in the formula we will get,

$\text{Volume of remaining solid}=\frac{2}{3}\pi*6^{2}*10$

$\text{Volume of remaining solid}=\frac{2}{3}\pi*36*10$

$\text{Volume of remaining solid}=2*\pi*12*10$

$\text{Volume of remaining solid}=240\pi$

$\text{Volume of remaining solid}=753.9822368615503772\approx 753.98$

Therefore, the volume of remaining solid is 753.98 cubic cm.

7 0

3 years ago

If DEF and MNO are two triangles such that DE=MN and EF=NO, which of the following would be sufficient to prove that triangles a

Delvig [45]

Answer:

D. ∠E ≅ ∠N

Step-by-step explanation:

The pair of sides meet at vertex E in ∆DEF and at vertex N in ∆MNO. Since the sides that make up angles E and N are shown congruent, it is sufficient to show ...

∠E ≅ ∠N

Then the SAS congruence postulate can be claimed.

<em>Additional comment</em>

The alternative is to show DF ≅ MO. That would allow you to claim SSS congruence. That is not an answer choice.

3 0

2 years ago

11/15-3/5 in fraction

Ket [755]

Answer:

l.c.m of 15 and 5 is 15

next step is to equalize the denominator with the l.c.m value

11/15 is the same (same denominator)

3/5 will change to 9/15 (multiply numerator and denominator by 3 to equalize)

Lastly, you just need to subtract

11/15 - 9/15

2/15

8 0

3 years ago

The length of a rectangle is twice its width. if the area of the rectangle is 200 yd2 , find its perimeter.

Lorico [155]

Width: x
Length: 2x
Area=x(2x)=2x^2
2x^2=200
x^2=100
x=10 or -10(reject as x>0) [width cannot be less than 0, doesnt make sense]
Perimeter
=x+x+2x+2x
=6x
=6(10)
=60 yd

7 0

2 years ago

Read 2 more answers

Solve for x:<br> 2(3x + 4) + 2 = 4 + 3x

Inga [223]

Answer:

-2 is the answer!!!!hope this helps

5 0

3 years ago