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zepelin [54]
2 years ago
9

Abby used the law of cosines for TriangleKMN to solve for k.

Mathematics
1 answer:
torisob [31]2 years ago
5 0

Answer:

  m∠K = 37° and n = 31

Step-by-step explanation:

A lot of math is about matching patterns. Here, the two patterns we want to match are different versions of the same Law of Cosines relation:

  • a² = b² +c² -2bc·cos(A)
  • k² = 31² +53² -2·31·53·cos(37°)

<h3>Comparison</h3>

Comparing the two equations, we note these correspondences:

  • a = k
  • b = 31
  • c = 53
  • A = 37°

Comparing these values to the given information, we see that ...

  • KN = c = 53 . . . . . . . . . . matching values 53
  • NM = a = k . . . . . . . . . . . matching values k
  • KM = b = n = 31 . . . . . . . matching values 31
  • ∠K = ∠A = 37° . . . . . . . matching side/angle names

Abby apparently knew that ∠K = 37° and n = 31.

__

<em>Additional comment</em>

Side and angle naming for the Law of Sines and the Law of Cosines are as follows. The vertices of the triangle are labeled with single upper-case letters. The side opposite is labeled with the same lower-case letter, or with the two vertices at either end.

Vertex and angle K are opposite side k, also called side NM in this triangle.

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3 years ago
Using the figure below, what is the value of y?
Bess [88]

Answer:

  C.  132

Step-by-step explanation:

We assume all angle measures are in degrees. The vertical angles are equal in measure, so ...

  4x = x +36

  3x = 36

  x = 12

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The value of y is 132.

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4 years ago
In the equation c*2=22 what is the next step in the equation solving sequence?
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3 years ago
The length (L) of a rectangle of fixed area varies inversely as the width (W).
Andreyy89

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2 years ago
Suppose the half-life of a material is 10 days. You have one 1 kg of the material today. How much of the material would you have
boyakko [2]

Answer:

0.50 kg of the material would be left after 10 days.

0.25 kg of the material would be left after 20 days.

Step-by-step explanation:

We have been given that the half-life of a material is 10 days. You have one 1 kg of the material today. We are asked to find the amount of material left after  10 days and 20 days, respectively.

We will use half life formula.

A=a\cdot(\frac{1}{2})^{\frac{t}{h}}, where,

A = Amount left after t units of time,

a = Initial amount,

t = Time,

h = Half-life.

A=1\text{ kg}\cdot(\frac{1}{2})^{\frac{10}{10}}

A=1\text{ kg}\cdot(\frac{1}{2})^{1}

A=1\text{ kg}\cdot\frac{1}{2}

A=0.5\text{ kg}

Therefore, amount of the material left after 10 days would be 0.5 kg.

A=1\text{ kg}\cdot(\frac{1}{2})^{\frac{20}{10}}

A=1\text{ kg}\cdot(\frac{1}{2})^{2}

A=1\text{ kg}\cdot\frac{1^2}{2^2}

A=1\text{ kg}\cdot\frac{1}{4}

A=0.25\text{ kg}

Therefore, amount of the material left after 20 days would be 0.25 kg.

6 0
3 years ago
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