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3 years ago

7

one sixth of the students in a school marching band are in the percussion section, the percussion section has 6 students, how ma

ny students are in the marching band?

2 answers:

gizmo_the_mogwai [7]3 years ago

8 0

Answer:Idk

Step-by-step explanation:

Igoryamba3 years ago

4 0

Answer:

36 students

Step-by-step explanation:

because if 1/6 is 6 students then you have to multiply the 6 by 6

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yKpoI14uk [10]

Answer:

<h2>6 unit cubes</h2>

Step-by-step explanation:

<em>Look at the picture.</em>

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3 years ago

Que costaría más comprar una vaca o comprar hamburguesas de 1€ hasta igualar a una vaca.

butalik [34]

Answer:

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2 years ago

You have 19 digs for your current volleyball season. There are 5 games left in the season. You want to break your previous recor

Charra [1.4K]

Answer:We must get more than 7 digs in the remaining 5 games to break record of 26.

Step-by-step explanation:

Given situation:- Let x represents the number of digs which we must get in the remaining 5 games to break your record ,and we have 19 digs for our current volleyball season.

To break previous record of 26 digs in a season, we have to get more than 26 i.e. 19 digs added to x must be greater than 26

⇒x+19 ≥ 26

⇒x≥26-19⇒x≥7

Therefore,we must get more than 7 digs in the remaining 5 games to break record of 26.

4 0

3 years ago

A ball of radius 15 has a round hole of radius 5 drilled through its center. Find the volume of the resulting solid. Hint: The u

OverLord2011 [107]

Answer:

The volume of the ball with the drilled hole is:

$\displaystyle\frac{8000\pi\sqrt{2}}{3}$

Step-by-step explanation:

See attached a sketch of the region that is revolved about the y-axis to produce the upper half of the ball. Notice the function y is the equation of a circle centered at the origin with radius 15:

$x^2+y^2=15^2\to y=\sqrt{225-x^2}$

Then we set the integral for the volume by using shell method:

$\displaystyle\int_5^{15}2\pi x\sqrt{225-x^2}dx$

That can be solved by substitution:

$u=225-x^2\to du=-2xdx$

The limits of integration also change:

For x=5: $u=225-5^2=200$

For x=15: $u=225-15^2=0$

So the integral becomes:

$\displaystyle -\int_{200}^{0}\pi \sqrt{u}du$

If we flip the limits we also get rid of the minus in front, and writing the root as an exponent we get:

$\displaystyle \int_{0}^{200}\pi u^{1/2}du$

Then applying the basic rule we get:

$\displaystyle\frac{2\pi}{3}u^{3/2}\Bigg|_0^{200}=\frac{2\pi(200\sqrt{200})}{3}=\frac{400\pi(10)\sqrt{2}}{3}=\frac{4000\pi\sqrt{2}}{3}$

Since that is just half of the solid, we multiply by 2 to get the complete volume:

$\displaystyle\frac{2\cdot4000\pi\sqrt{2}}{3}$

$=\displaystyle\frac{8000\pi\sqrt{2}}{3}$

5 0

3 years ago

Help asap im in a quiz!

goldfiish [28.3K]

The answer should be b

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2 years ago