10. Find the slope of the line.

A. 3

B.

C. –3

D.

   The answer is either letter A or letter D

0

Hint 1: Find the direction cosines of AB and BC. Then take their dot product. This will give the cosine of the angle between AB and BC.

Hint 2: The Direction Cosines of PQ, P(x1,y1,z1) and Q(x2,y2,z2), are (±x2−x1(x2−x1)2+(y2−y1)2+(z2−z1)2√,±y2−y1(x2−x1)2+(y2−y1)2+(z2−z1)2√,±z2−z1(x2−x1)2+(y2−y1)2+(z2−z1)2√)

Edit: The Dot Product works iff the lines intersect.

B is the intersect point , doesn't matter if it's in 2D or 3D – Ali Apr 28 '18 at 5:52

0

Lines in dimension three do not have to intersect. We find their direction vectors to see if they are parallel.

Note that

AB→=<x2−x1,y2−y1,z2−z1>

and

BC→=<x3−x2,y3−y2,z3−z2>

These vectors are the direction vectors are your lines.

The dot product

AB→.BC→

along with the norms of these vectors determine the angle between the lines if they intersect at all.

cosθ=AB→.BC→||AB||.||BC||

Step-by-step explanation:    This is for the picture

Center:  (0, 0)

Width:  10

Angle:  0 rad

Height:  6.8

Opacity:  1

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

17.08

correct me if i am wrong.