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Nookie1986 [14]
2 years ago
7

M(-5,9) and N(-2,7) P(-3,-7) and Q(3,-5

Mathematics
1 answer:
jek_recluse [69]2 years ago
7 0

Answer:

M=(4,112)=(4,5.5) A

Step-by-step explanation:

The midpoint for two points P=(px,py) and Q=(qx,qy) is M=(px+qx2,py+qy2).

We have that px=3, py=2, qx=5, qy=9.

Thus, M=(3+52,2+92)=(4,112).

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NNADVOKAT [17]

Answer:

SA=234.8\ yd^2

Step-by-step explanation:

we know that

The surface area of the regular pyramid is equal to the area of the triangular base plus the area of its three triangular lateral faces

step 1

Find the area of the triangular base

we know that

The triangular base is an equilateral triangle

so

The area applying the law of sines is equal to

A=\frac{1}{2}(14^2)sin(60^o)

A=\frac{1}{2}(196)\frac{\sqrt{3}}{2}

A=49\sqrt{3}=84.87\ yd^2            

step 2

Find the area of its three triangular lateral faces

A=3[\frac{1}{2}bh]

we have

b=14\ yd

Find the height of triangles

Applying the Pythagorean Theorem

10^2=(14/2)^2+h^2

solve for h

100=49+h^2

h^2=100-49

h=\sqrt{51}\ yd

substitute

A=3[\frac{1}{2}(14)\sqrt{51}]      

A=149.97\ yd^2

step 3

Find the surface area

Adds the areas

SA=84.87+149.97=234.84\ yd^2

Round to the nearest tenth

SA=234.8\ yd^2

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