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3 years ago

I xant solve for the value of K

Mathematics

1 answer:

exis [7]3 years ago

4 0

K = -3x^2/5 -5x -3 + px/5. I hope that helps.

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Prove that<br>{(tanθ+sinθ)^2-(tanθ-sinθ)^2}^2 =16(tanθ+sinθ)(tanθ-sinθ)

USPshnik [31]

First, expand the terms inside the bracket you will get

$(( \tan {}^{2} (x) + 2 \tan(x) \sin(x) + \sin {}^{2} (x) - ( \tan {}^{2} (x) - 2 \tan(x) + \sin {}^{2} (x) ) {}^{2} = 16( \tan(x) + \sin(x) )( \tan(x) - \sin(x) )$

$( 4 \tan(x) \sin(x) ) {}^{2} = 16( \tan(x) + \sin(x) )( \tan(x) - \sin(x) )$

$16 \tan {}^{2} (x) \sin {}^{2} (x) = 16( \tan(x) + \sin(x) )( \tan(x) - \sin(x) )$

$16 \tan {}^{2} (x) (1 - \cos {}^{2} (x) ) = 16 (\tan(x) + \sin(x) )( \tan(x) - \sin(x) )$

$16( \tan {}^{2} (x) - \frac{ \sin {}^{2} (x) \cos {}^{2} ( {x}^{} ) }{ \cos {}^{2} (x) }$

$16( \tan {}^{2} (x) - \sin {}^{2} (x) ) = 16( \tan(x) + \sin(x) )( \tan(x) - \sin(x) )$

$16( \tan(x) + \sin(x) )( \tan(x) - \sin(x) = 16( \tan(x) + \sin(x) )( \tan(x) - \sin(x) )$

5 0

2 years ago

HELP WITH THIS QUESTION PLEASE!!

svetlana [45]

m<PQS = m<SQR

4y - 10 = 2y + 10

2y = 20

y = 10

m<PQR = m<PQS + m<SQR

= 4y - 10 + 2y + 10

= 6y

= 6•10

= 60

3 0

3 years ago

2 fifth= Ten blank help

Paha777 [63]

2/5 = x/10

2/5 x 2 = 4/10

x= 4

2/5 = 4/10

6 0

2 years ago

What is 75% of 180?

OverLord2011 [107]

Hello There!

75% of 180 is 135.

Converting Percent To Decimal.

p = 75%/100 = 0.75

Y = 0.75 * 180

Y = 135

3 0

3 years ago

Read 2 more answers

Which of these examples is an error of addition by one?

N76 [4]

Answer:

The correct answer is:

A. 27,465 +52,534 = 79,899

Step-by-step explanation:

Given are the options for addition

We have to check each addition one by one

So,

A. 27,465 +52,534 = 79,899

This option is the example of error of addition by one. The answer should have been 79,999. Hence there is an error in addition ..

Hence, option a: 27,465 +52,534 = 79,899 is the correct answer ..

4 0

3 years ago

Read 2 more answers