Correct answers only ! a comedian needs up to ten volunteers for a show. he needs no fewer than four female volunteers. let x represent the number of female volunteers and y represent the number of male volunteers. select all inequalities that model this situation.

this one is kind of tricky!

So we can select anything that fits within the confines of the requirements of the comedian. He wants at least 4 females and could use anywhere up to 10 if there are no males. Using the bare minimum of 4 females, he could use anywhere up to 6 boys. If there are 3 or less females, we assume there is no show. But there doesn't have to be any boys for the show as long as he has at least 4 females.

Okay. I think I've got it. Now let's look at each example and see if it fits the parameters above using x for females and y for males.

1. x (girls) is greater or less than 4. YES this fits.

2. y (boys) are greater than 0. Well, they could be, but remember we can do the show with NO boys so this doesn't work. NO.

3. x plus y is less than or equal to 10. That is true so YES.

4. y is greater or equal to 0. YES. (this is how #2 should have looked)

5. x is greater than 4. Well, it could be. But x can also be equal to 4 so this doesn't work. NO

6. x plus y is less than 10. It could be, but this doesn't allow for the possiblity that they could be equal to 10. So NO this does not work.

We have: 1, 3, 4 as correct models of the situation.

step-by-step explanation:

assume that you have 3 different side lengths.   apply the pythagorean theorem here:   square each of the two shortest sides; call them a and b, so that their squares are a² and b².   also square the longest side, calling the result c².

now, if a² + b² = c² is true, the triangle is a right triangle; otherwise it is not a right triangle.