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kicyunya [14]
3 years ago
12

The winner of a basketball contest is the individual who scores the highest percentage of free throws. The top three finishers a

re contestants A, B, and C. Contestant A made 85% of her free throws, contestant B made 0.88 of his free throws, and contestant C made 2125
of his free throws. Graph an inequality that represents the possible percentages (in decimal form) of free throws made by the other contestants in the competition. (Assume there are no ties.)
Mathematics
1 answer:
Kamila [148]3 years ago
3 0

Answer:

<h2>Less than 0.84</h2>

Step-by-step explanation:

  1. Convert all numbers to decimals
  • 85% --> 0.85
  • 0.88 --> 0.88
  • 21/25 --> 0.84

Since these are the top three, the rest of the competitors are less than 0.84.

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What is the area of the shaded section?
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How many triangles does a=6 b=10 A=33° create?
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Answer:

2 triangles are possible.

Step-by-step explanation:

Given

a=6

b=10

\angleA=33°

To find:

Number of triangles possible ?

Solution:

First of all, let us use the <em>sine rule</em>:

As per Sine Rule:

\dfrac{a}{sinA}=\dfrac{b}{sinB}

And let us find the angle B.

\dfrac{6}{sin33}=\dfrac{10}{sinB}\\sinB = \dfrac{10}{6}\times sin33\\B =sin^{-1}(1.67 \times 0.545)\\B =sin^{-1}(0.9095) =65.44^\circ

This value is in the 1st quadrant i.e. acute angle.

One more value for B is possible in the 2nd quadrant i.e. obtuse angle which is: 180 - 65.44 = 114.56^\circ

For the value of \angle B = 65.44^\circ, let us find \angle C:

\angle A+\angle B+\angle C = 180^\circ\\\Rightarrow 33+65.44+\angle C = 180\\\Rightarrow \angle C = 180-98.44 = 81.56^\circ

Let us find side c using sine rule again:

\dfrac{6}{sin33}=\dfrac{c}{sin81.56^\circ}\\\Rightarrow c  = 11.02 \times sin81.56^\circ = 10.89

So, one possible triangle is:

a = 6, b = 10, c = 10.89

\angleA=33°, \angleA=65.44°, \angleC=81.56°

For the value of \angle B =114.56^\circ, let us find \angle C:

\angle A+\angle B+\angle C = 180^\circ\\\Rightarrow 33+114.56+\angle C = 180\\\Rightarrow \angle C = 180-147.56 = 32.44^\circ

Let us find side c using sine rule again:

\dfrac{6}{sin33}=\dfrac{c}{sin32.44^\circ}\\\Rightarrow c  = 11.02 \times sin32.44^\circ = 5.91

So, second possible triangle is:

a = 6, b = 10, c = 5.91

\angleA=33°, \angleA=114.56°, \angleC=32.44°

So, answer is : 2 triangles are possible.

8 0
2 years ago
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