MATHEMATICS
HIGH SCHOOL

Answer: The probability to choose letter n is . After drawing letter n there left 25 letters and the probability to choose the letter w is . After drawing two letters n and w there left 24 letters and then the probability to choose letter b is . After drawing three letters n, w and b there left 23 letters, so the probability to choose letter t is .

Using the product rule for probabilities, you can obtain that .

Using the product rule for probabilities, you can obtain that .

MIDDLE SCHOOL

There are 318 fiction books in the class library.the number of nonfiction books is 47 less than the number of fiction books about how many nonfiction books are there in the class library?explain

318 books are fiction, 47 books are nonfiction. 318-47=271

COLLEGE

I just need help for 3 questions! Write an expression for the perimeter of the triangle shown below:

A triangle is shown with side lengths labeled 2.3x plus 14, 2x, and negative 0.2x plus 15.

4.1x + 29

4.1x − 29

4.5x + 29

4.5x − 29

Which expression is equivalent to 1/5m − 20?

1/5(m − 4)

1/5(m − 100)

5(m − 4)

5(m − 100)

Write an expression that is equivalent to 3/5 (3y + 15).

three over fivey + 9

three over five y + 15

nine over fivey + 9

nine over fivey + 15

Answers:

1. a) 4.1x + 29

2. b) 1/5(m – 100)

3. c) nine over five y + 9 (9/5y + 9)

Explanations:

1. The formula for perimeter of a triangle is P = side a + side b + side c, so with these values it is:

2.3x + 14 + 2x – .2x + 15

Combine like terms:

4.1x + 29

2. Test the options to see which ones work out to the answer. You know that 1/5 of 100 = 20, so you need one with a 100 and a 1/5 in it. The only possible answer is:

1/5(m – 100)

3. Distribute the 3/5:

3/5 • 3y + 3/5 • 15 = 9/5y + 45/5 = 9/5y + 15

1. a) 4.1x + 29

2. b) 1/5(m – 100)

3. c) nine over five y + 9 (9/5y + 9)

Explanations:

1. The formula for perimeter of a triangle is P = side a + side b + side c, so with these values it is:

2.3x + 14 + 2x – .2x + 15

Combine like terms:

4.1x + 29

2. Test the options to see which ones work out to the answer. You know that 1/5 of 100 = 20, so you need one with a 100 and a 1/5 in it. The only possible answer is:

1/5(m – 100)

3. Distribute the 3/5:

3/5 • 3y + 3/5 • 15 = 9/5y + 45/5 = 9/5y + 15

Answer:

Answers:

1. a) 4.1x + 29

2. b) 1/5(m – 100)

3. c) nine over five y + 9 (9/5y + 9)

Explanations:

1. The formula for perimeter of a triangle is P = side a + side b + side c, so with these values it is:

2.3x + 14 + 2x – .2x + 15

Combine like terms:

4.1x + 29

2. Test the options to see which ones work out to the answer. You know that 1/5 of 100 = 20, so you need one with a 100 and a 1/5 in it. The only possible answer is:

1/5(m – 100)

3. Distribute the 3/5:

3/5 • 3y + 3/5 • 15 = 9/5y + 45/5 = 9/5y + 15

Step-by-step explanation:

MIDDLE SCHOOL

What is the solution to the inequality? - 9x – 15<3

x > 2

x <-2

x < 2

x >-2

Answer:

x > -2

Step-by-step explanation:

First: - 9x – 15 < 3, you would want to take the 15 and add from each side.

Now it looks like: - 9x < 18

Second: You want to take the -9 and divide it from 18

Now it looks like: x < -2

Third: Chnage the less than sign into a greather than sign because we are dealing with a negative #

Now it looks like: x > -2

HIGH SCHOOL

Evaluate the discriminant of each equation. Tell how many solutions each equation has and whether the solutions are real or imaginary x^2 + 4x + 5 = 0

x^2 - 4x - 5 = 0

4x^2 + 20x + 25 = 0

All these equations are in the form of ax^2 + bx + c = 0, where a, b, and c are some numbers. the discriminants of equations like this are equal to b^2 - 4ac. if the discriminant is negative, there are two imaginary solutions. if the discriminant is positive, there are two real solutions. if the discriminant is 0, there is one real solution.

x^2 + 4x + 5 = 0

b^2 - 4ac

4^2 - 4(1)(5)

16-20

-4, two imaginary solutions.

x^2 - 4x - 5 = 0

b^2 - 4ac

(-4)^2 - 4(1)(-5)

16 + 20

36, two real solutions.

4x^2 + 20x + 25 = 0

b^2 - 4ac

20^2 - 4(4)(25)

400 - 400

0, one real solution.

x^2 + 4x + 5 = 0

b^2 - 4ac

4^2 - 4(1)(5)

16-20

-4, two imaginary solutions.

x^2 - 4x - 5 = 0

b^2 - 4ac

(-4)^2 - 4(1)(-5)

16 + 20

36, two real solutions.

4x^2 + 20x + 25 = 0

b^2 - 4ac

20^2 - 4(4)(25)

400 - 400

0, one real solution.