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The largest value of c such that -2 is in the range of .x^2 + 3x + c - 7x + x^2 is 0.

\(c=-2x^2+4x-2\\ \frac{dc}{dx}=-4x+4=0\\ x=1\\ c=-2\cdot 1^2+4\cdot 1-2\\ \color{blue}c=0\)

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The sum of all possible values of x is 7.

The answer is 48.

Suppose f(0) is positive.  Then f(x) must remain above x for all x, but this is not possible at the end because f(x) must be squeezed between 0 and 1.  So f(a) = a for some a.

bro you can't compute a domain with a floor function

The 182nd bead is red.

it was actually [200,680) but thank you for trying to help me :)

I solved this before!  The answer is [200,550).

if x^2 + 9 = 10, then x would have been +- 1, both less than 5. ELIMINATED

if 3x - 8 = 10, then x = 6, also ELIMINATED.

But i think there is a typo, do you mean "f(x) = x^2 + 9 if x > -5", then the first case would work, and the answer would hence be 0.

The pythagorean theorem tells us that the distance from P to Q is sqrt(64+9) = sqrt(73). A = pi r^2. A = 73pi.

\(\frac{\ln{b}}{\ln{a}}\)

If "the square" refers to the large one, then 50

If it's the small one, then 100.

A counter-example is given by f(x) = x*sin(1/x).

Can you show us the steps?

@You-Know-Who, i don't think the conjugate root theorem applies here, since the coefficients are complex too.

72*2/3=48

48+53.43= 101.43 

sean has $101.43

x^2=x*x

x^3=x*x*x

x^3=5x^2

x*x*x=5*x*x

divide by x^2

x=5

1 x 1 triangles: 3

1 x 2 triangles: 2

1 x 3 triangles: 1

2 x 1 triangles: 3

2 x 2 triangles: 2

2 x 3 triangles: 1

3 x 1 triangles: 3

3 x 2 triangles: 2

3 x 3 triangles: 1 

So the sum is \(\color{brown}\boxed{18}\)

The possible values of t are 5 and 6.

P(z) = z^2 + bz + c = (z - (-7 + 2i))(z - (-1 - 2i))

P(1) = b + c + 1 = (7 - 2i)(2 + 2i)

14 + 14i - 4i + 4 - 1 

17 + 10i.

60 70+50=120 180-120=60.    so its 60