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1. You just need to add up the x^3 terms, which is just x^3+5x^3 = 6x^3, which means the answer is 6.

2. Since both of the polynomials are of degree 4, the largest possible value of the degree of f(x)+a*g(x) is still 4, since multiplying a polynomial by a constant doesn't change its degree.

3. To find the smallest possible degree, try to cancel out the x^4 term. To do that, set b = -1/2. That will make the x^4 term of g(x) equal to -1, which cancels out with the x^4 term of f(x), but notice that the x^2 terms also cancel out. That means the smallest possible degree is 1. 

4. since both of the polynomials are of degree 4, the degree of f(x)*g(x) is just 4+4=8.

5. very similar to question 2, the answer is 11.

6. hint: this is the same question as solving this equation:

28-5+2t=0

1. False, because it is not symmetrical across the y axis.

2. Can't tell, because although it seems like it's false, it could be true if you zoom out enough.

3. True, because the graph goes down and then up at 2 points.

4. Can't tell, because although it seems like it's false, it could be true if you zoom out enough (similar to statement 2).

A) geometric sequence, and the sum is just \(-\frac{512+256+128+\dots+2+1}{512} = \boxed{-\frac{1023}{512}}\).

(notice that 1023 is one smaller than 1024, which is 2^10)

B) Idk if my solution is even close to optimal, but here it is:

Let the common difference be x. Then, 

\((35-6x)+(35-5x)+(35-4x)+(35-3x)+(35-2x)+(35-x)+(35)+(35+x)=220\\ 8(35) - 20x=220\\ -20x=-60\\ \boxed{x=3}\)

The common difference is therefore equal to 3

The first term is just \(35-6(3)=\boxed{17}\)

varvaax lmao I'm not that good :)

idk what happened to my answer, but I meant \(3\leq x \leq 7\)

anyway, for the second part of the problem, the angle opposite of x will be acute only if \(3^2+5^2>x^2\rightarrow\sqrt{34}>x\). For integer values, that means that it must be smaller or equal to 5.

So that means that if x is 3, 4, or 5, it will make the angle opposite of x be acute, but that doesn't mean that the triangle itself is acute. If x = 4, then it will for a 3-4-5 triangle, which is right. That means that only 3 and 5 works, so the answer is \(\boxed{2}\)

By the way, if you want to manually check if a triangle is obtuse or not, and you're given sides x, y, and z, then all of these these inequalities must hold true:

\(x^2+y^2>z^2\\ y^2+z^2>x^2\\ x^2+z^2>y^2\)

If x adds 4 when y adds 12, then if x adds 1, y adds 12/4 = 3. 

Therefore, from -1 to 0, x adds 1, and y adds 3, so the answer is -7+3 = -4

Let x = CA. By the triangle inequality, \(5-3 . There are  \(\boxed{5}\) integer solutions.

The green line has a length of 2, the blue line has a length of 7, and the red line has a length of 15.

Since the green line is parallel to the blue line, the 2 right triangles shown above must be similar to each other.

That means the smaller red length is just \(\frac{2}{9}\cdot15 = \boxed{\frac{10}{3}}\), which is the answer.

Guest was right for the first part, but the number of ways to pass through the point (2, 2, 2) is not \(\frac{6!}{2!2!2!}\), it's \((\frac{6!}{2!2!2!})^2 = 8100\), so the answer is \(\boxed{26550}\)

Hint: set up a system of equations:
\(2\pi r h=3.5\\ \pi r^2h=3.5\)

Divide the second equation by the first one:

\(\frac{\pi r^2h}{2\pi r h}=1\\ \)

then, cancel out like terms. 

Can you take it from there?