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Maksim231197 [3]
3 years ago
9

Which pair of segments in the figures below are congruent?

Mathematics
2 answers:
SSSSS [86.1K]3 years ago
6 0

Answer:

segment BC & segment PN

Step-by-step explanation:

just did the quiz

oksian1 [2.3K]3 years ago
3 0

Answer:

Segment B C and Segment P N

Step-by-step explanation:

we know that

Line segments are congruent if they have the same length

In this problem

we have

Segment L.M ≅ Segment F.A

Segment M.N ≅ Segment A.B

Segment P.N ≅ Segment B.C

Segment P.J ≅ Segment C.D

Segment J.K ≅ Segment E.D

Segment K.K ≅ Segment F.E

therefore

The answer is

Segment B C and Segment P N

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2/6, 5/12, 3/7, and 4/10. List least to most
olya-2409 [2.1K]
Turn them into percentages:
2/6=1/3=33.333%
5/12=42%
3/7=43%
4/10=2/5=40%

Now put them in order :)

2/6,4/10,5/12,3/7
7 0
2 years ago
Can you simplify (9c^8)^1/2
Tamiku [17]

Answer: 9c8

 ———

  2

Step-by-step explanation:

32c8

Simplify   32c8———

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7 0
3 years ago
Pls help me with this one <br><br><br>1,2,3,4​
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Answer:

Option 2.

Explanation:

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5 0
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Step-by-step explanation:

4 0
3 years ago
find the zeros of 2x^2 - 16x + 27 using the quadratic formula. be sure to simplify the expression. Can someone please show me ho
____ [38]
For a quadratic of the form f(x)=ax^2+bx+c, we have the quadratic formula 
x=\dfrac{-b \pm \sqrt{b^2 -4ac}  }{2a},
where a is the coefficient (number before the variable) of the squared term, b is the coefficient of the linear term, and c is the constant term.

So, given 2x^2-16x+27=0, we can get that a=2, \ b=-16, and c=27. We substitute these numbers into the quadratic formula above.

x=\dfrac{-(-16) \pm \sqrt{(-16)^2 -4(2)(27)} }{2(2)}

x=\dfrac{16 \pm \sqrt{(256 -216)} }{4}

x=\dfrac{16 \pm \sqrt{40} }{4}

x=\dfrac{16 \pm 2\sqrt{10} }{4}

x=4+ \frac{\sqrt{10}}{2}, \ x=4- \frac{\sqrt{10}}{2}

This is our final answer.

If you've never seen the quadratic formula, you can derive it by completing the square for the general form of a quadratic. Note that the \pm symbol (read: plus or minus) represents the two possible distinct solutions, except for zero under the radical, which gives only one solution.
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3 years ago
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