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antoniya [11.8K]
1 year ago
10

based only on the information given in the diagram, which congruence theorems or postulates could be given as reasons why triang

le hel is congruent to triangle pme
Mathematics
2 answers:
nydimaria [60]1 year ago
5 0

Answer:

i would pick

HL ,HA, ASA

Step-by-step explanation:

quester [9]1 year ago
3 0
<span>Hi :) Firstly we look at the side that is actually the same. How many sides are there that are the same? Next, we note the number of angles that are the same. </span>
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Answer:

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How many solutions does the equation sin(5x) = 1/2 have on the interval (0, 2PI]
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Answer:

Step-by-step explanation:

Given the equation

Sin(5x) = ½

5x = arcSin(½)

5x = 30°

Then,

The general formula for sin is

5θ = n180 + (-1)ⁿθ

Divide through by 5

θ = n•36 + (-1)ⁿ30/5

θ = 36n + (-1)ⁿ6

The range of the solution is

0<θ<2π I.e 0<θ<360

First solution

When n = 0

θ = 36n + (-1)ⁿθ

θ = 0×36 + (-1)^0×6

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When n = 1

θ = 36n + (-1)ⁿ6

θ = 36-6

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When n = 2

θ = 36n + (-1)ⁿ6

θ = 36×2 + 6

θ = 78°

When n =3

θ = 36n + (-1)ⁿ6

θ = 36×3 - 6

θ = 102°

When n=4

θ = 36n + (-1)ⁿ6

θ = 36×4 + 6

θ = 150

When n =5

θ = 36n + (-1)ⁿ6

θ = 36×5 - 6

θ = 174°

When n = 6

θ = 36n+ (-1)ⁿ6

θ = 36×6 + 6

θ = 222°

When n = 7

θ = 36n + (-1)ⁿ6

θ = 36×7 - 6

θ = 246°

When n =8

θ = 36n + (-1)ⁿ6

θ = 36×8 + 6

θ = 294°

When n =9

θ = 36n + (-1)ⁿ6

θ = 36×9 - 6

θ = 318°

When n =10

θ = 36n + (-1)ⁿ6

θ = 36×10 + 6

θ = 366°

When n = 10 is out of range of θ

Then, the solution is from n =0 to n=9

So the equation have 10 solutions in the range 0<θ<2π

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1 year ago
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Answer:

48

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