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ANTONII [103]
3 years ago
7

Y the

Mathematics
1 answer:
Eduardwww [97]3 years ago
6 0

Answer: b and d

Step-by-step explanation:

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You are graphing Square ABCDABCDA, B, C, D in the coordinate plane. The following are three of the vertices of the square: A(4,
Kitty [74]

Answer:

D(4,-3)

Step-by-step explanation:

Given three of the vertices of the square: A(4, -7), B(8, -7),C(8, -3)

Let the coordinate of the fourth vertex be D(x,y).

We know that diagonals of a square are perpendicular bisector. So, the midpoint of both diagonals is the same.

The diagonals are BD and AC

Midpoint of BD = Midpoint of AC

\left(\dfrac{8+x}{2},\dfrac{-7+y}{2}\right) =\left(\dfrac{4+8}{2},\dfrac{-7+(-3)}{2}\right)\\ \left(\dfrac{8+x}{2},\dfrac{y-7}{2}\right) =\left(\dfrac{12}{2},\dfrac{-10}{2}\right)\\ \left(\dfrac{8+x}{2},\dfrac{y-7}{2}\right) =\left(6,-5\right)\\$Therefore$:\\\dfrac{8+x}{2}=6\\8+x=12\\x=12-8\\x=4\\$Similarly$\\\dfrac{y-7}{2}=-5\\y-7=-5*2\\y-7=-10\\y=-10+7=-3

The coordinates of the fourth vertex is D(4,-3)

3 0
3 years ago
Craig won first prize in Mrs. Short's class
Bogdan [553]

Answer:

1/11

Step-by-step explanation:

There are 7 gift card options so that makes it 7/77 which simplified is 1/11

6 0
3 years ago
Read 2 more answers
1. Use the given key to answer the questions.
Eduardwww [97]

Answer:

please where is the diagram

8 0
3 years ago
Need help on this question
sergey [27]
Total Area: T.A.=2*Ab+Al

Area of the base: Ab=p*K

Semi-perimeter of the base: p
p=P/2
Perimeter of the base: P=20
p=P/2=20/2→p=10
Ab=p*k=10*K→Ab=10K

Lateral Area of the prism: Al
Al=P*h
Height of the prism: h=6
Al=P*h=20*6→
Al=120

T.A.=2*Ab+Al
T.A.=2*(10K)+120
T.A.=20K+120
T.A.=120+20K

Answer: (120+20K)



7 0
3 years ago
3m+8=15<br><img src="https://tex.z-dn.net/?f=3m%20%2B%208%20%3D%2015" id="TexFormula1" title="3m + 8 = 15" alt="3m + 8 = 15" ali
Umnica [9.8K]

Answer:

m = \frac{7}{3}

Step-by-step explanation:

Given

3m + 8 = 15 ( subtract 8 from both sides )

3m = 7 ( divide both sides by 3 )

m = \frac{7}{3}

8 0
3 years ago
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