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2 years ago

In circle F with m∠EFG=54m∠EFG=54 and EF=19

Mathematics

1 answer:

elena-14-01-66 [18.8K]2 years ago

3 0

Answer:

17.907082 unit

Step-by-step explanation:

According to the Question,

Given, A circle with centre F, ∠EFG=54 and EF=19 .

length of arc EG = Radius(EF) × ∠EFG(in Radian)

We Know, 1 degree = 0.0174533 Radian
54 degree = 0.942478 Radian

length of arc EG = 19 x 0.942478 ⇔ 17.907082 unit

(For Diagram please find in attachment)

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Cos^2x+cos^2(120°+x)+cos^2(120°-x)<br>i need this asap. pls help me

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Answer:

$\frac{3}{2}$

Step-by-step explanation:

Using the addition formulae for cosine

cos(x ± y) = cosxcosy ∓ sinxsiny

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cos(120 + x) = cos120cosx - sin120sinx

= - cos60cosx - sin60sinx

= - $\frac{1}{2}$ cosx - $\frac{\sqrt{3} }{2}$ sinx

squaring to obtain cos² (120 + x)

= $\frac{1}{4}$ cos²x + $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x

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cos(120 - x) = cos120cosx + sin120sinx

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squaring to obtain cos²(120 - x)

= $\frac{1}{4}$ cos²x - $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x

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Putting it all together

cos²x + $\frac{1}{4}$ cos²x + $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x + $\frac{1}{4}$ cos²x - $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x

= cos²x + $\frac{1}{2}$ cos²x + $\frac{3}{2}$ sin²x

= $\frac{3}{2}$ cos²x + $\frac{3}{2}$ sin²x

= $\frac{3}{2}$ (cos²x + sin²x) = $\frac{3}{2}$

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Luke buys shares for £15.26 each they decrease in value by 3.5% work out the new share price

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Step-by-step explanation:

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A bowl contains 20 red balls and 20 blue balls. A man selects balls at random without looking at them. How many balls must he se

mario62 [17]

Answer:

we will select 9 balls to be sure of having at least five balls of the same color.

Step-by-step explanation:

Given that:

The bowl contains (20 red + 20 blue) balls = 40 balls

If we are to select 5 balls out of the 40 balls.

Let R = red and B = blue

The combinations can be:

5R + 0 B

4R + 1 B

3R + 2 B

1R + 4 B

0 R + 5 B

From above, we will notice that there are valid proof that there are 5 balls of the same color.

If we select six balls, we have:

= (6R + 0 B ), (5R + 1 B ), (4R + 2 B ), (3R + 3 B ), (2R + 4 B ), (1R + 5 B ), (0R + 6 B )

If we select seven balls, we have:

= (7R + 0 B ), (6R + 1 B ), (5R + 2 B ), (4R + 3 B ), (3R + 4 B ), (2R + 5 B ), (1R + 6 B ), (0R + 6 B ).

If we select eight balls, we have:

= (8R + 0 B ), (7R + 1 B ), (6R + 2 B ), (5R + 3 B ), (4R + 4 B ), (3R + 5 B ), (2R + 6 B ), (1R + 7 B ), (0R + 8B)

Now, selecting nine balls, we have the following combinations:

= (9R + 0 B ), (8R + 1 B ), (7R + 2 B ), (6R + 3 B ), (5R + 4 B ), (4R + 5 B ), (3R + 6 B ), (2R + 7 B ), (1R + 8B), (0R + 9B).

Thus, we can see that all the combinations comprise of 5 balls with at least one color.

Therefore, we will select 9 balls to be sure of having at least five balls of the same color.

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