CubeyThePenguin

Here is one example:

You have $20 and oranges cost $3 each. At the counter, you decide to order a magazine worth $2. How many oranges can you buy?

If x is the number of oranges, then 3x + 2 <= 20.

This is by no means the only example that can be used. A Google search would likely reveal more than enough examples to reference.

Expand by distribution.

-2(8 - 7x)

= -2(8) - 2(-7x)

= -16 + 14x

2/3x - 9 = -13

2/3x = -4

2 = -4(3x)

2 = -12x

x = -1/6

0.579 * 104 = 60.216

(x - 3)(x^2 + 4x + 4) = x - 3 + x^3 - 7x + 8

x^3 + x^2 - 8x - 12 = x^3 - 6x + 5

x^2 - 8x - 12 = -6x + 5

x^2 - 2x - 17 = 0

From there, use the quadratic formula and add the roots to get the answer.

(x + 2)(3x^2 + x + 5)

= (x)(3x^2 + x + 5) + 2(3x^2 + x + 5)

= (3x^3 + x^2 + 5x) + (6x^2 + 2x + 10)

= 3x^3 + 7x^2 + 7x + 10

I'm assuming the question is a typo and you want to find Ax^3 + Bx^2 + Cx + D. It shouldn't be too hard to match up the coefficients to find the answer.

Let the original number be called x. We can rewrite the original problem as:

(five times a number) is divided by (7 more than that number) = -8

5x/(x + 7) = -8

5x = -8(x + 7)

5x = -8x - 56

13x = -56

x = -56/13

The general term is \(\binom{4}{n}(6x)^{n}(\frac{3}{x})^{4-n}\). For the 6x and 3/x terms to balance out, n must equal 2.

\(\binom{4}{2} (6x)^2 (\frac{3}{x})^2 = 6 \cdot 36x^2 \cdot \frac{9}{x^2} = \boxed{1944}.\)

P(thunderstorms on Tuesday and Wednesday)

= P(thunderstorms on Tuesday) * P(thunderstorms on Wednesday)

= 40% * 65%

= 26%

Separate this into two geometric sequences.

(1 + 1/6 + 1/36 + ...) + (1/2 + 1/12 + ...)

= 1/(1 - 1/6) + (1/2)/(1 - 1/6)

= 6/5 + (1/2)*(6/5)

= 9/5

The discriminant of this quadratic is

\(b^2 - 4ac = (-m+2)^2 - 4(m)(m-8) = -3m^2 + 28m + 4\)

For starters, the discriminant has to be nonnegative, which limits our values of m to the solutions of -3m^2 + 28m + 4 >= 0. So, m can be 0, 1, 2, 3, or 4. From there, it's easy to try each individual value until you find an equation with integer roots. 

Let P(x) = x^5 - x^4 + x^3 - px^2 + qx + 14. From the remainder theorem, if P(x) is divisible by x+2, then P(-2) = 0. (The same holds true for x-1.)

So,

P(-2) = -32 - 16 - 8 + 4p - 2q + 14 = 0

P(1) = 1 - 1 + 1 - p + q + 14 = 0

4p - 2q = -42

-p + q = -15

Solving this system of equations will yield the answer.

For each ball, there are three choices of which box to put it in. So, the answer is 4^3 = 64.

This question has now been answered:

https://web2.0calc.com/questions/volume-of-a-cube_1#r1

1. 20, 40, 60, ... , 500, 520

2. 80, 100, ..., 1000

3. 32, 40, ...,  72, 80

4. a) 2, 4, 6, ..., 122, 124

b) 3, 6, 9, ..., 120, 123

c) 6, 12, ..., 114, 120

d) combine the list of numbers in a) and b)

5. 102, 104, ..., 998, 1000

b) 108, 117, ..., 999

c) 108, 126, ..., 990

d) combine the list of numbers in a) and b)

Unfortunately, we can't help you if there is no question posted.

question reposted: JD is repainting some of the walls in the classrooms at Golder. One wall has a width that is 4 feet less than its length. The wall has an area of 140 square feet. What is the length of the wall? 

Let the length be L. We can write an equation for the area:

L(L - 4) = 140

L^2 - 4L - 140 = 0

(L - 14)(L + 10) = 0

L = 14 or L = -10

Length can't be negative, so the wall's length is 14 feet.

For a), you could write about either the first cube or the second cube. 

example:

a) What is the volume of the first cube?

b) 7^3 = 343 cm^3.