Triangles k l p and q m n are shown. triangle q m n is slightly higher than triangle k l p and side q m connects to side k p. point m is at the midpoint of k p. sides k l and q n are congruent. angles k l p and q n m are congruent. angles k p l and q m n are both right angles. is there a series of rigid transformations that could map triangleklp to triangleqnm? if so, which transformations? no, triangleklp and triangleqnm are congruent but triangleklp cannot be mapped to triangleqnm using a series rigid transformations. no, triangleklp and triangleqnm are not congruent. yes, triangleklp can be reflected across the line containing kp and then translated so that p is mapped to m. yes, triangleklp can be rotated about p and then translated so that l is mapped to n.

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}

spinning a 1, 2, or 3. option d.

step-by-step explanation: when spinning a number graph up to 4, and aiming for any number less than 3, that automatically eliminates the number 4. now you have two numbers to spin on that are less than 3, but the number 3 is included in this question because it's no greater of an number than itself. i hope this you : )

answer: yes i agree that you should not put it on/p>

step-by-step explanation: the reason for this is that it wastes other peoples time and also make them fail as well and that is never good.

the answer is 11.2 (i know i’m late)