BuilderBoi

You're welcome!!

o yea, forgot 4 was composite lol 

and also i already accounted for feb 29 and feb 31 

Here's my attempt:

Let the function be in the form \(ax^3 + bx^2 + cx + d = y\). Also note that \(f(1) = 9\) means that \((1,9)\) is a point on this graph. Subbing this in gives us \(a + b + c + d = 9\)

With stars and bars, we use 3 bars, so it is \({9 + 3 \choose 3} = 220\). But, this accounts for \(a = 0\), which we don't want; it has to be a cubic polynomial. 

If we "remove a", there are 9 stars, but only 2 boxes, so there are \({9 + 2 \choose 2} = 55\) ways for \(a = 0\).

So, there are \(220 - 55 = \color{brown}\boxed{165}\) polynomials. 

There are 5 "prime months" (2, 3, 5, 7, 11)

For the dates, there are 12 (2, 3, 4, 5, 7, 11, 13, 17, 19, 23, 29, 31)

So there are \(5 \times 12 = 60\), but we overcount for 2/29, 2/31, and 11/31, so there are \(60 - 3 = \color{brown}\boxed{57}\) "prime dates".

Inequality 1: \(3n + 8< 30\)

Subtract 8 from both sides: \(3n <22\)

Divide by 3: \(n < {22 \over 3}\)

Inequality 2: \(2n - 9 > -4\)

Add 9 to both sides: \(2n > 5\)

Divide both sides by 2: \(n > {5 \over 2}\)

So our range becomes \({5 \over 2} < n < {22 \over 3}\)

The integers that work are 3, 4, 5, 6, and 7.

So the answer is \(3 \times 4 \times 5 \times 6 \times 7 = \color{brown}\boxed{2520}\)

here: https://web2.0calc.com/questions/help-due-in-1-hour#r1

Multiplying the entire equation by \(x^3 - x^2 - 21x + 45\): \(1 = A(x-3)(x-3) + B(x+5)(x-3) + C(x-5)\)

Simplify the right-hand side: \(1 = A(x^2 - 6x + 9) + B(x^2 + 2x - 15) + C(x-5)\)

Equating terms to get the following system: \(Ax^2 + Bx^2 = 0\)

                                                                     \(-6Ax + 2Bx + Cx = 0\)

                                                                    \(9A - 15B - 5C = 1\)

Removing the x's from the system gives us a simpler system: \(A + B = 0\)

                                                                                                   \(-6A + 2B + C = 0\)

                                                                                                   \(9A - 15B - 5C = 1\)

Now, just solve for a. 

:)

The formula for the radius of the sphere is \(4 \pi r^2\). We know that this is equal to \(196 \pi\). Setting these equal and solving for r, we find that \(r = 7\), meaning the circumference is \(2 \pi r = \color{brown}\boxed{14 \pi}\)

You can't hear other guesses tho...

that part alone makes it soo hard 

Let the amount of money Jeri owns be \(j\). Off the bat, we know \(j \geq 200\).

Represent the first part as \({2 \over 3}(j-80) + 80\). The second part represents how much money Jeri has in her right pocket, and the latter represents the amount of money in her left pocket. The second part is just \(j - 200\).

So we have the inequality: \({2 \over 3}(j-80) + 80 > j-200\)

Solving, we find \(j<680\), but because we know \(j \geq 200\), we have \(200 \leq j < 680\), which equals \(\color{brown}\boxed{[200, 680)}\) in interval notation.