GYanggg

Never seen a problem like this before.

Well done!

I'm done for today, I will continue on the rest of your problem tommorow. 

I know how to do them but I'm just too tired. 

We first need to simplify both inequalities. 

4x + 3 < 25

Subtracting 3 from both sides:

4x < 22

Dividing both sides by 4:

x < 11/2

-7n + 5 < 24

Adding both sides by 7n and subtracting both sides by 24 (to avoid dividing by a negative number):

7n > -19

Dividing both sides by 7:

n > -19/7

n must be greater than -2 and -5/7 but also smaller than 5 and 1/2

The integers in between are: 

-2, -1, 0, 1, 2, 3, 4, and 5

There are 8 integers 

Hi gueesstt, to make this a square of a bunomial, we must take the square of one half of the coefficient of the term where the degree of x is 1. 

So the coefficient of the term where the degree of x is 1, is 5/3.

We must take the square of one half of that.

First, one half of 5/3 is 5/6.

Then we take the square of that, which is 25/36. 

That is the number that makes the quadratic a square of a binomial. 

Since all four girls are next to each other, we can treat them like one person. 

So it is basically a six person permutation.

\(6!=720\)

There are 720 ways to arrange the boys and girls, but we can also change the order the 4 girls sit in. 

\(4!=24\)

In total, there are \(24\cdot720=17280.\) ways to arrange everyone in a row. 

I hope this helped,

Gavin.

You didn’t specify what we need to find, but assuming you want BC, here is my solution:

Using the Pythagorean Thereom, BD^2 = AB^2 - AD^2 = 13^2 - 5^2 = 12^2 

BD = 12

Using the same method, BC^2 = DC^2 - BD^2 = 20^2 - 12^2

You take the square root of that and you have your answer.

Alright, here we go:

For two number's product to be positive, those numbers must be both positive or both negative, but cannot be zero.

This is because zero is neither positive nor negative, and 0 * n = 0. 

The first step is to find the total number of combination: (a,b).

This can be done by:

\((1-(-3)+1)\cdot(4-(-2)+1) =5\cdot7=35.\)

There are a total of 35 different combinations. 

Out of those combinations there are:

( -3 , -2 ) ; ( -3 , -1 ) ; ( -2 , -2 ) ; ( -2 , -1 ) ; ( -1 , -2 ) ; ( -1 , -1 )

( 1 , 1 ) ; ( 1 , 2 ) ; ( 1 , 3 ) ; ( 1 , 4 )

10 that work. 

The probabilty of this happening is:

\(\boxed{\frac{10}{35}=\frac{2}{7}}\)

I hope this helped, 

Gavin.

Hi tertre,

Since line AB is parallel to line CD, angle AXE is equal to CYX.

This means we can set angle AXE = angle CYX = x

Since AXE is 108º less than 3 times CYX, we have this:

\(x+108=3x\)

Solving for x, we get:

\(x=54\)

We know that AXE and CYX = 54º

Since AXE and BXY are congruent angles, we know that

\(BXY=54º\)

I hope this helps, 

Gavin

I do not see the figure to the right, may you please repost?

Hello Guest!

Alright, first we need to take care of the positions with special conditions. 

The lineman has a special condition, which only three of them are strong enough to do. 

There are \(\binom{3}{1}\)ways to choose a lineman. 

The rest doesn't need any special players. 

So the total amount of ways is:

\(\binom{3}{1}+\binom{9}{1}+\binom{8}{1}+\binom{7}{1}=3+9+8+7=27\)

I'm pretty sure that is the final answer. 

I hope this helps, 

Gavin.