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GalinKa [24]
3 years ago
9

Consider △LNM. Triangle L M N is shown. Angle N M L is a right angle. Which statements are true for triangle LNM? Check all that

apply. The side opposite ∠L is NM. The side opposite ∠N is ML. The hypotenuse is NM. The hypotenuse is LN. The side adjacent ∠L is NM. The side adjacent ∠N is ML.

Mathematics
1 answer:
horsena [70]3 years ago
3 0

Answer:

  • The side opposite ∠L is NM.
  • The side opposite ∠N is ML.
  • The hypotenuse is LN.

Step-by-step explanation:

Even a crude drawing can be helpful as you sort this out. (See attached.)

In general, the side opposite an angle <u>will not</u> have the letter of the angle in its name. Similarly, a side adjacent to the angle <u>will</u> have the angle letter in its name.

The statements that apply are ...

  • The side opposite ∠L is NM.
  • The side opposite ∠N is ML.
  • The hypotenuse is LN.   (opposite right angle M)

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Answer:

  y +5 = -1/3(x -3)

Step-by-step explanation:

The slope of the given line is 3. The slope of the perpendicular line is the negative reciprocal of that, -1/3. Then the point-slope equation is ...

  y -y1 = m(x -x1) . . . . line with slope m through point (x1, y1)

  y -(-5) = -1/3(x -3)

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4x = 8x − 1<br><br> x = three fourths<br> x = 1<br> x = 3<br> x = 6
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Isolate the variable by subtracting 8x from both sides to get -4x = -1. multiply both sides by -1 to get 4x = 1. divide both sides by 4 to get x = 1/4.
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Step-by-step explanation:

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Lin has a scale model of a modern train. The model is created at a scale of 1 to 48
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Step-by-step explanation:

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A team of 10 players is to be selected from a class of 6 girls and 7 boys. Match each scenario to its probability. You have to d
tankabanditka [31]
The selection of r objects out of n is done in

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C(13,10)= \frac{13!}{3!(10)!}= \frac{13*12*11*10!}{3*2*1*10!}= \frac{13*12*11}{3*2}=  286
thus, the sample space of the experiment is 286

A. 
<span>"The probability that a randomly chosen team includes all 6 girls in the class."

total number of group of 10 which include all girls is C(7, 4), because the girls are fixed, and the remaining 4 is to be completed from the 7 boys, which can be done in C(7, 4) many ways.


</span>C(7, 4)= \frac{7!}{4!3!}= \frac{7*6*5*4!}{4!*3*2*1}= \frac{7*6*5}{3*2}=35
<span>
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B.
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</span>C(6, 3)= \frac{6!}{3!3!}= \frac{6*5*4*3!}{3!3!}= \frac{6*5*4}{3*2*1}=20
<span>
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C.
"</span>The probability that a randomly chosen team has either 4 or 6 boys.<span>"

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this was already calculated in part A, it is </span>0.12.
<span>
case 2, the team has 6 boys and 4 girls.

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<span>
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D.
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