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lidiya [134]
3 years ago
8

Find the midpoint of the segment connecting (4,7) to (-2,3).

Mathematics
1 answer:
KonstantinChe [14]3 years ago
8 0
If you put 2 points on a graph paper and connect them, you can find the midpoint pretty easily
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maria has 15 fewer apps on her phone than orlando. if the total number of apps on both phones is 29, how many apps are on each p
butalik [34]

Answer: 14

Step-by-step explanation:

29-15=14

Let me know if I helped :D

3 0
3 years ago
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Write and solve an algebraic equation to find x triangle<br> (2x) (5x+12) (3X-32)
denis23 [38]

Answer:

30x squared 2 minus 248x squared 2 minus 768x

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3 years ago
Can someone plz help me with this one problem plz I’m being timed!!!
katrin [286]
1 = 2
2= 8
3 = 14
4 = 20

Cant see the graph, but whichever goes through these points. Use Desmos to graph the function if you need further help.
4 0
3 years ago
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10 + 25
mezya [45]

Answer:

(x+5)^2

Step-by-step explanation:

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4 0
3 years ago
Use the squared identities to simplify 2sin^2xsin^2x​
nataly862011 [7]

Answer:

option A  

Step-by-step explanation:

cos(4x) = cos(2x+2x) = cos(2x)*cos(2x) - sin(2x)*sin(2x)

cos(4x) = cos^2(2x) - sin^2(2x)

cos(4x) = cos^2(2x) - cos^2(2x) - 1     (using cos^2(2x) + sin^2(2x) = 1)

cos(4x) = 2cos^2(2x) - 1   (eq. 1)

cos(2x) = cos(x+x) = cos(x)*cos(x) - sin(x)*sin(x)

cos(2x) = cos^2(x) - sin^2(x)

cos(2x) = 1 - sin^2(x) - sin^2(x)     (using cos^2(x) + sin^2(x) = 1)

cos(2x) =  1 - 2sin^2(x)    (eq. 2)

Replacing eq. 2 into eq. 1:

cos(4x) = 2[1 - 2sin^2(x)]^2 - 1

cos(4x) = 2[1 - 4sin^2(x) + 4sin^4(x)] - 1

cos(4x) = 1 - 8sin^2(x) + 8sin^4(x)

cos(4x) - 1 + 8sin^2(x) = 8sin^4(x)

cos(4x)/4 - 1/4 + 2sin^2(x) = 2sin^4(x)

2sin^2(x)sin^2(x)​ = 2sin^4(x) = cos(4x)/4 - 1/4 + 2sin^2(x)

Using eq. 2:

2sin^2(x)sin^2(x)​ = cos(4x)/4 - 1/4 + 1 - cos(2x)

2sin^2(x)sin^2(x)​ = cos(4x)/4 + 3/4 - 4cos(2x)/4

2sin^2(x)sin^2(x)​ = [3 + cos(4x) - 4cos(2x)]/4

7 0
3 years ago
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