CubeyThePenguin

The inradius can be found using the formula A = rs, where A is the triangle's area, r is the inradius, and s is the triangle's semiperimeter.

\(s= \frac{29+29+40}{2} = 49\)

\(A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(49)(20)(20)(9)} = 420\)

\(\rightarrow r = \frac{A}{s} = \frac{420}{49} = \boxed{\frac{60}{7}}\)

We can make a Venn Diagram:

[baseball || both || football] neither

[     37     ||   x    ||      26   ]    24

37 + 26 - x + 24 = 75

x = 12

1. either not on the football team OR on the baseball team

The set of people on the baseball team is a subset of the people not on the football team.

probability = (37 + 24)/75 = 61/75

2. at least on the baseball team OR not on the football team

This problem has the same answer as #1. (The wording is kind of confusing; I don't think "at least... or" has any difference from "either... or".)

3. either on the football team OR not on the baseball team

The set of people on the football team is a subset of the people not on the baseball team.

probability = (26 + 24)/75 = 50/75 = 2/3

4. at least on the football team OR not on the baseball team

This problem has the same answer as #3.

5. neither on the football nor the baseball team

This information was given in the problem.

probability = 24/75 = 8/25

There are 52 - 3 = 49 cards remaining, since the 4 of hearts, 8 of clubs, and 9 of hearts have already been drawn. We need the next card drawn to be higher than a 9. Out of the possible values, the next card could be a 10, J, Q, or K (assuming face cards are greater than 10).

In that case, there are 4 * 4 = 16 possible cards that would give a winning hand for a probability of 16/49.

a) Let the number of permits printed be "x."

\(500 + 2x = 6x\)

b) We will solve the equation from part a).

Subtract 2x from both sides:

\(500 = 4x\)

Divide both sides by 4:

\(125 = x\)

Mercury High School needs to sell 125 parking permits for the average cost of a permit to be $6.

We can make a Venn Diagram:

[ring || both || necklace] neither

[20%|| x%   ||    30%    ]   60%

a) probability that the student is wearing a ring or a necklace:

1 - 60% = 40%

b) probability that the student is wearing a ring and a necklace:

20 + 30 - x + 60 = 100

110 - x = 100

x = 10%

In general, the probability of getting "x" at-bats is:

\(\binom{25}{x}(0.3089)^x(1-0.3089)^{25-x}\)

To find the probability that the player gets greater than 9 hits, we can either substitute x = 9, 10, ..., 25 and add the results OR substitute x = 0, 1, ..., 8, add the results, and subtract that from 1. The rest of the problem boils down to plugging numbers into the calculator.

the roots are x = -7 and x = 8:

\((x+7)(x - 8)\)

\(= x^2 + 7x - 8x - 56\)

\(= x^2 - x - 56\)

(This expression can be expanded by using FOIL: First, Outer, Inner, Last.)

Because our equation's x^2 coefficient is already equal to 1, our answer is x^2 - x - 56 = 0.

\(\sqrt{60x} \cdot \sqrt{12x} \cdot \sqrt{63x} \cdot \sqrt{42x}\)

\(= \sqrt{60x \cdot 12x \cdot 63x \cdot 42x}\)

\(= \sqrt{(2^2 \cdot 3 \cdot 5) \cdot (2^2 \cdot 3) \cdot (3^2 \cdot 7) \cdot (2 \cdot 3 \cdot 7) \cdot x^4}\)

\(= \sqrt{2^5 \cdot 3^5 \cdot 5 \cdot 7^2 \cdot x^4}\)

\(= 2^2 \cdot 3^2 \cdot 7 \cdot x^2 \cdot \sqrt{2 \cdot 3 \cdot 5}\)

\(= \boxed{252x^2 \sqrt{30}}\)

tl;dr: split each number into their prime factors, multiply them all, then move the perfect squares outside the square root

\((2y - 7)^5\)

\(= \binom{5}{0}(2y)^5 + \binom{5}{1}(2y)^4(-7)^1 + \binom{5}{2}(2y)^3(-7)^2 + \binom{5}{3}(2y)^2(-7)^3 + \binom{5}{4}(2y)(-7)^4 + \binom{5}{5}(-7)^5\)

We only need the coefficient of the y^4 term:

\(\binom{5}{1}(2y)^4(-7)^1\)

\(= 5 \cdot 16y^4 \cdot (-7) = \boxed{-560y^4}\)

Therefore, the coefficient of y^4 is -560.

For further information, study the binomial theorem. Here is a link to get started:

https://courses.lumenlearning.com/boundless-algebra/chapter/the-binomial-theorem/

2x(x + 10) = -50

2x^2 + 20x = -50

2x^2 + 20x + 50 = 0

Using VIeta's formulas, the sum of roots of the quadratic ax^2 + bx + c = 0 is -b/a.

-b/a = -20/2 = -10. 

(As ElectricPavlov has solved this question, the two roots are -5 and -5, which sum to -10 as expected.)