Awesomeguy

Options a and b are eliminated because we don't know anything about f(x) when x is in range -1 < x < 2.

Looking at c and d, we try to see which one makes sense. Because we know in this range of -2 < x < 1, f(x) will be negative, we can rewrite c and d.

c) = |f(x)| = f(x) because the -1 * -1 = 1

d) |f(x)| = -f(x) since f(x) in range -2 < x < 1 is negative.

Therefore the answer is c).

Its just the multiples of 3 from the smallest possible sum to the largest. 

From 3 + 6 + 9 = 18, to 36 + 39 + 42 = 117.

This is from the 6th multiple of 3, to the 39th multiple of 3.

6 - 5 = 1, 39 - 5 = 34

There are 34 distinct sums.

The formula of a midpoint of two points is just the average of the x, the average of the y.

$(\frac{(6+x)}{2},\frac{(8+y)}{2}) = (-1, -1) \\ 6+x = -2 \\ 8+y = -2 \\ x = -8 \\ y = -10$

-8 + -10 = -18

That means the roots are -3 and 8. Putting it in factored form,

$(x+3)(x-8) \\ x^2 - 5x -24$

oh oops wow i mess that up lol

nice solution!!

ohhhhhhh oops lol

If the first term is x, second is x + y, and so on,

The fifth term would be $x + 4y = 9$ and 32nd would be $x + 31y = -84$. Subtract and solve for x and y,

$x + 4y = 9 \\ x+ 31y = -84 \\ 27y = -93 \\ y = -\frac{31}{9} \\ x + 4(-\frac{31}{9})=9 \\ x - 13 \frac{7}{9} = 9 \\ x = 22 \frac{7}{9}$

Equation for 23nd term would be $x + 22y$.

$22 \frac{7}{9} + 4(-\frac{31}{9}) \\ 22 \frac{7}{9} + - 13\frac{7}{9} = 9$

So the 23rd term is 9.

Multiply all 4 equations together,

$m^3a^3t^3h^3= 27 \\ math=3$

Divide by each equation,

$h = 24 \\ t = \frac{3}{32} \\ a = 9 \\ m = \frac{4}{27}$

Equation: y = -3x + 91
$91 + 17 = 108 \\ 108 / 3 = 36 \\ y = -3(36) + 91 = -108 + 91 = -17$

So 35 terms will come before -17 appers.

Similar problem: https://web2.0calc.com/questions/trapezoid-amp-circle

Equation of line,

$y = 2x + b$

Plugging in (8, 0),

$0 = 2(8) + b \\ 0 = 16 + b \\ b = -16$

So the y-intercept has y-coordinate -16, so it is at point, (0, -16)