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KIM [24]
3 years ago
9

Four football fans took turns driving the distance from new York to Oklahoma to see a big game

Mathematics
1 answer:
Snezhnost [94]3 years ago
4 0
I'm sorry but what's the full question?
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What is the perimeter of this triangle? 15 cm 5 cm 25 cm 10 cm
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55

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2 years ago
Write out the sample space for the given experiment. Use the letter R to indicate red, G to indicate green, and B to indicate bl
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Answer for the question:

Write out the sample space for the given experiment. Use the letter R to indicate red, G to indicate green, and B to indicate blue. A die shows 33 different colors on it. Give the sample space for the next 22 rolls.

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3 0
3 years ago
PLEASE PLEASE ANSWER!
muminat

The area of the shaded region is $8(\pi \ -\sqrt{3})\ \text{cm}^2.

Solution:

Given radius = 4 cm

Diameter = 2 × 4 = 8 cm

Let us first find the area of the semi-circle.

Area of the semi-circle = \frac{1}{2}\times \pi r^2

                                      $=\frac{1}{2}\times \pi\times 4^2

                                      $=\frac{1}{2}\times \pi\times 16

Area of the semi-circle = $8\pi cm²

Angle in a semi-circle is always 90º.

∠C = 90°

So, ABC is a right angled triangle.

Using Pythagoras theorem, we can find base of the triangle.

AC^2+BC^2=AB^2

AC^2+4^2=8^2

AC^2=64-16

AC^2=48

AC=4\sqrt{3} cm

Base of the triangle ABC = 4\sqrt{3} cm

Height of the triangle = 4 cm

Area of the triangle ABC = \frac{1}{2}\times b \times h

                                          $=\frac{1}{2}\times 4\sqrt{3}  \times 4

Area of the triangle ABC =  8\sqrt{3} cm²

Area of the shaded region

                   = Area of the semi-circle – Area of the triangle ABC

                   = $8\pi \ \text{cm}^2-8\sqrt{3}\ \text{cm}^2

                   = $8(\pi \ -\sqrt{3})\ \text{cm}^2

Hence the area of the shaded region is $8(\pi \ -\sqrt{3})\ \text{cm}^2.

3 0
2 years ago
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