1. what are the endpoint coordinates for the midsegment of △jkl that is parallel to jl?
enter your answer, as a decimal or whole number, in the boxes.
( ) ( ), and ( ) ( )
2. in parallelogram abcd , diagonals ac and bd intersect at point e, be=2x2−3x , and de=x2+10 .
what is bd ?
enter your answer in the box.
units
3.jklm is a parallelogram.
what is the measure of ∠klj?
enter your answer in the box.

Part 1) and

Part 2)

Part 3) m∠KLJ=

Step-by-step explanation:

Part 1)

we have

Find the coordinates of the midpoint JK

the x-coordinate of the midpoint JK is equal to

the y-coordinate of the midpoint JK is equal to

The midpoint JK is the point

Find the coordinates of the midpoint LK

the x-coordinate of the midpoint LK is equal to

the y-coordinate of the midpoint LK is equal to

The midpoint LK is the point

The answer part 1) is

the endpoint coordinates for the midsegment of △JKL that is parallel to JL are the points   and

Part 2)

we know that

The diagonals bisect the parallelogram into two congruent triangles

In the parallelogram ABCD

substitute the values

Solve the quadratic equation

The formula to solve a quadratic equation of the form is equal to

in this problem we have

so

substitute in the formula

Find the value of BD

Substitute the value of

Part 3)

we know that

The diagonals bisect the parallelogram into two congruent triangles

In the parallelogram JKLM

m∠KLJ=m∠MLJ

we have that

m∠MLJ=

therefore

m∠KLJ=

there were a lot of them but the first one to pop up on a google search was bts

hello,

as we don't know if g(x) is continue,

answer b

f(x)=2x²-8x+1=2(x²-4x+4)+1-8=2(x-2)²-7

vertex is (2,-7) ( a minimum since coefficient of x² is > 0)