Help can someone help me find the m

Mathematics

2 answers:

Sliva [168]1 year ago

6 0

It’s going to be 140, if you need to show work it’s 180-130=50 then 180-90=90 then 90+50=140

7nadin3 [17]1 year ago

5 0

∠AED + ∠AEB = 180° (linear pairs)

∠AED + 90° = 180°

∠AED = 90°

Again,

∠DEC + ∠BEC = 180° (linear pairs)

∠DEC + 130° = 180°

∠DEC = 180° - 130°

∠DEC = 50°

Now,

∠AEC = ∠DEC + ∠AED = 90° + 50° = 140°

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algol [13]

The answer is b
The way I do it is the area of the front back times 2 which is 2x4x2=16
Then top and bottom times 2 which is 4x8x2=64
Then left and right which is 2x8x2=32 then you add them all together so 16+64+32=112
Hope this helped :)

8 0

4 years ago

Point M is the midpoint of PQ. The coordinates of P and M are given below.

Zepler [3.9K]

The coordinates of Q are (-1,6)

Further Explanation:

Given points are:

$(x_m, y_m) = (5, -2)\\(x_p,y_p) = (11, -10)$

The formula for mid-pint is:

$(x_m, y_m) = (\frac{x_p+x_q}{2} , \frac{y_p+y_q}{2})\\From\ this\ we\ get\\x_m = \frac{x_p+x_q}{2}\\Putting\ values\\5 =\frac{11+x_q}{2}\\5*2 = 11+x_q\\10 = 11+x_q\\10-11 = x_q\\x_q = -1\\AND\\y_m = \frac{y_p+y_q}{2}\\-2 = \frac{-10+y_q}{2}\\-2*2 = -10+y_q\\-4 = -10+y_q\\-4+10 = y_q\\y_q = 6$