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2 years ago

The question is -x^2+18x-75.

1 answer:

5 0

When you use -b/2a you plug in a and b. in this case it would be -18/-2 which coms out as 9. then you plug 9 into x. -9^2+18(9)-75 which comes out as -12. Your vertex is (9,-12) your Axis of semmetry would be 9.

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Find the measure of angle x in the figure below: A.) 35° B.) 48° C.) 69° D.) 78°

prisoha [69]

Answer:

x = 48

Step-by-step explanation:

First find y

52+y+59 = 180 since they from a straight line

y = 180-59-52

y = 69

Then we know the angles of a triangle add to 180

x+y+63 = 180

x+69+63 = 180

x = 180-69-63

x =48

3 0

3 years ago

Read 2 more answers

Two points on L1 and two points on L2 are given. Use the slope formula to determine if lines L1 and L2 are parallel, perpendicul

vlada-n [284]

Answer:

The lines L1 and L2 neither parallel nor perpendicular

Step-by-step explanation:

* Lets revise how to find a slope of a line

- If a line passes through points (x1 , y1) and (x2 , y2), then the slope

of the line is $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

- Parallel lines have same slopes

- Perpendicular lines have additive, multiplicative slopes

( the product of their slopes is -1)

* Lets solve the problem

∵ L1 passes through the point (1 , 10) and (-1 , 7)

- Let (1 , 10) is (x1 , y1) and (-1 , 7) is (x2 , y2)

∴ x1 = 1 , x2 = -1 and y1 = 10 , y2 = 7

∴ The slope of L1 is $m1 = \frac{7-10}{-1-1}=\frac{-3}{-2}=\frac{3}{2}$

∵ L2 passes through the point (0 , 3) and (1 , 5)

- Let (0 , 3) is (x1 , y1) and (1 , 5) is (x2 , y2)

∴ x1 = 0 , x2 = 1 and y1 = 3 , y2 = 5

∴ The slope of L2 is $m2=\frac{5-3}{1-0}=\frac{2}{1}=2$

∵ m1 = 3/2 and m2 = 2

- The two lines have different slopes and their product not equal -1

∴ The lines L1 and L2 neither parallel nor perpendicular

7 0

2 years ago

In which quadrant is the number –14 – 5i located on the complex plane?

loris [4]

Answer:

The complex number $r = -14 - i\,5$ belongs to the third quadrant of the complex plane.

Step-by-step explanation:

Let be $r = -14 - i\,5$ . In the complex plane, if $\Re (r) < 0$ (real component) and $\Im (r) < 0$ (imaginary component), the number belongs to the third quadrant of the complex plane. The complex number $r = -14 - i\,5$ belongs to the third quadrant of the complex plane.