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Mars2501 [29]
2 years ago
9

In a school of 1200 pupils 45% are boys how many boys are there in the school

Mathematics
2 answers:
mrs_skeptik [129]2 years ago
8 0
540 of the students at the school are boys
45%/100%=x/1200
1200×45=5400
5400÷100=540
aleksklad [387]2 years ago
7 0
Hello,

Shall we begin?

45% = 45/100 = 0,45

= 1.200 * 0,45
= 540

<span>Answers: 540 boys</span>
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3 years ago
Differentiate <br> \[\ln \sin^2 x\]
iragen [17]
Y = ln {[sin(x)]^2}

use the chain rule

y' = 1 /[sin(x)]^2 * 2sin(x)*cos(x) = 2cos(x) / sin(x) = 2 cot(x)

Answer: 2 cot(x)


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3 years ago
The movie starts at 3:15 p.m. and ends at 5:08 p.m. How long is the movie?
Katena32 [7]

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1 hour and 53 minutes

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4 0
3 years ago
What is .16 repeating as a simplified fraction?
Readme [11.4K]
I'm assuming this is 0.1666666...

Which in that case is 1/6.

(Just from memory)

To do 0.161616161616...

First, assign 0.161616... a variable, say it's x.

Now say 100x, what is it.

16.1616161616...

Now subtract x from 100x.

16.161616161616... - 0.16161616...

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8 0
3 years ago
Evaluate the six trigonometric functions of the angle 0<br><br><br><br>is right<br>​
antiseptic1488 [7]

Answer:

1. Sine θ = 1/3

2. Cos θ = 2√2 / 3

3. Tan θ = √2 / 4

4. Cosec θ = 3

5. Sec θ = 3√2 / 4

6. Cot θ = 2√2

Step-by-step explanation:

We'll begin by determining the adjacent. This can be obtained as follow:

Hypothenus (Hypo) = 9

Opposite (Opp) = 3

Adjacent (Adj) =?

Hypo² = Adj² + Opp²

9² = Adj² + 3²

81 = Adj² + 9

81 – 9 = Adj²

72 = Adj²

Take the square root of both side

Adj = √72

Adj = 6√2

Finally, we shall determine six trigonometric functions of the angle θ. This Can be obtained as follow:

1. Determination of Sine θ

Hypothenus = 9

Opposite = 3

Sine θ =?

Sine θ = Opposite / Hypothenus

Sine θ = 3/9

Sine θ = 1/3

2. Determination of Cos θ

Adjacent = 6√2

Hypothenus = 9

Cos θ =?

Cos θ = Adjacent / Hypothenus

Cos θ = 6√2 / 9

Cos θ = 2√2 / 3

3. Determination of Tan θ

Opposite = 3

Adjacent = 6√2

Tan θ =?

Tan θ = Opposite / Adjacent

Tan θ = 3 / 6√2

Tan θ = 1 / 2√2

Rationalise

(1 / 2√2) × (2√2 /2√2)

= 2√2 / 4×2

Tan θ = √2 / 4

4. Determination of Cosec θ

Sine θ = 1/3

Cosec θ =?

Cosec θ = 1 / Sine θ

Cosec θ = 1 ÷ 1/3

Cosec θ = 1 × 3/1

Cosec θ = 3

5. Determination of sec θ

Cos θ = 2√2 / 3

Sec θ =?

Sec θ = 1 / Cos θ

Sec θ = 1 ÷ 2√2 / 3

Sec θ = 1 × 3 / 2√2

Sec θ = 3 / 2√2

Rationalise

= (3 / 2√2) × (2√2 / 2√2)

= 3 × 2√2 / 4×2

Sec θ = 3√2 / 4

6. Determination of Cot θ

Tan θ = √2 / 4

Cot θ =?

Cot θ = 1 / Tan θ

Cot θ = 1 ÷ √2 / 4

Cot θ = 1 × 4 / √2

Cot θ = 4 / √2

Rationalise

= (4 / √2) × (√2 / √2)

= 4√2 / 2

Cot θ = 2√2

8 0
3 years ago
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