MN has endpoints at M(2,5) and N(-4, 3). In what quadrant is the midpoint
of MN?
A Quadrant |
B Quadrant II
C Quadrant IIl

quad 2

Step-by-step explanation:

Midpoint = (x1 + x2)/2,(y1 + y2)/2

x1 = 2

x2 = - 4

y1 = 5

y2 = 3

midpoint = (2 - 4)/2,(5 + 3)/2

midpoint = -2/2, 8/2

midpoint = -1,4

-1 for x puts you in quad 2 or 3

4 for y puts you in quad 2 because you are going up.

The final answer is in quad 2

the answer is d

step-by-step explanation:

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}