Jk has endpoints at j(7, 10) and k(3, 2). find the midpoint m of jk.

Answers

37/3

step-by-step explanation:

step 1: simplify both sides of the equation.

7x+8(x+

1

4

)=3(6x−9)−8

7x+(8)(x)+(8)(

1

4

)=(3)(6x)+(3)(−9)+−8(distribute)

7x+8x+2=18x+−27+−8

(7x+8x)+(2)=(18x)+(−27+−8)(combine like terms)

15x+2=18x+−35

15x+2=18x−35

tep 2: subtract 18x from both sides.

15x+2−18x=18x−35−18x

−3x+2=−35

step 3: subtract 2 from both sides.

−3x+2−2=−35−2

−3x=−37

step 4: divide both sides by -3.

−3x

−3

=

−37

−3

x=

37

3

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation:

J (7, 10)
K (3, 2)

Formula for midpoint:
\frac{x1 + x2}{2}
x - coordinate for M:
\frac{7 + 3}{2}  = 5
y - coordinate for M:
\frac{10 + 2}{2}  = 6
Therefore, M (5,6).

Hope this helps :)


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