GYanggg

I already answered, but here it is:

The cylindrical storage container as a volume of:

\(5^2\pi\cdot7\approx550\)

Since 550 is greater than 500.

To answer your question,

The cat litter will fit, since the volume of the cat liter is less than the storage. 

The cylindrical storage container as a volume of:

\(5^2\pi\cdot7\approx550\)

Since 550 is greater than 500.

To answer your question,

The cat litter will fit, since the volume of the cat liter is less than the storage. 

On the left side of the equation, we can tell that that we can easily factor it. 

It is a perfect cube. 

We use \((x-y)^3=x^3-3x^2y+3xy^2-y^3\)

Then we have:

\((\frac{x}{3}-3)^3=64\)

\(\frac{x}{3}-3=4 \)

x = 21

I hope this helped,

Gavin

Hi supermanaccz!

So from 100 to 200, we first need to determine what numbers are bad. 

From 100 to 159, they are all good, since we can have 1:00 and 1:59.

But from 160 to 199, they are all bad. There is no 1:60, since that will be 2 o'clock. 

There is no 1: 99.

So we need to calculate how many numbers from 160 - 199 are equal to 3 mod 4.

The smallest number that is from 160 - 199 that is equal to 3 mod 4 is 163.

3 / 4 = 0 R 3

163 / 4 = 40 R 3

The largest is 199. 

199 / 4 = 49 R 3

To calculate how many numbers we want from 163 to 199, we do:

\((199-163)/3+1=\boxed{13}\)

That will be your final answer .

I hope this helped, 

gavin.

You can do adults = x

and say students = 45 - x

this is just easier

i hope this helps

gavin

88.

There are a total of:

\(\binom{15}{4}\)

ways to choose the debaters. 

This is equal to 1365.

Since we need those four people, with only one way to choose them.

The chance this will happen is 1/1365. 

87. 

Since there are three digits and each digit has 10 options,

there are a total of:

10^3 ways of identifying students. 

The first number can be a 0, since it is not supposed to be a number, just an identification. 

10^3 = 1000

That is you final answer. 

Oh wow, well done. 

I did not think of this method, writing an equation. 

I though we make it a whole circle and manipulate it somehow, but I just couldn't find the area of the rectangle. 

Using the rule about the sides of a triangle, where the sum of any two sides must be greather than the third. 

We have these inequalities:

7 + 3 > x

7 + x > 3

x + 3 > 7

Simplifying these inequalities, we have:

x < 10

x > -4

x > 4

The possible integer lengths are:

5, 6, 7, 8, and 9.

I hope this helped, 

Gavin.

We use Heron's formula. 

\(\sqrt{s(s-a)(s-b)(s-c)}\)

Where s is half of the perimeter.

We can first solve for s, which is:

\((5+5+6)\div2=8\)

Plugging our values into the formula, we have:

\(\sqrt{8(8-5)(8-5)(8-6)}\)

The final answer is 12, 

I hope this helped, 

Gavin