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2 years ago

Factorise the following quadratic expressions; 6a-18a21x -3x²1/16 - p²25v²-1

Mathematics

1 answer:

Lubov Fominskaja [6]2 years ago

3 0

Answer:

− 12 a

3 x ( 7 − x )

( 1 4 + p ) ( 1 4 − p )

( 5 v + 1 ) ( 5 v − 1 )

Step-by-step explanation:

I think these are the right answers

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What’s the absolute value of 3

RSB [31]

Answer: 3

Explanation: Absolute value means <em>distance from zero</em> on a number line. So far the absolute value of 3, we know that 3 is 3 units from zero on a number line so the absolute value of 3 is 3.

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3 years ago

On simplifying (-342-8) - (-5/2 + 1)

icang [17]

Answer:

-348.33

Step-by-step explanation:

Use PEMDAS as your guide.

(-342-8) - (-5/2+1)

PEMDAS

parenthesis

-350 - (-5/3)

PEMDAS

Dvisision

-350- (-1.67)

= - 348.33

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3 years ago

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morpeh [17]

I got 2.17 but I'm not sure that's right.

$180 - (71 + 90) = (6x + 4)$

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3 years ago

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valkas [14]

Answer:

I use schoology too.

Step-by-step explanation:

Just copy and paste it and look it up. Internet always helps.

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3 years ago

Verify that this trigonometric equation is an identity?

Vlada [557]

$\cot x\sec^4 x=\cot x+2\tan x+\tan^3x\\\\L=\dfrac{\cos x}{\sin x}\cdot\dfrac{1}{\cos^4x}=\dfrac{1}{\sin x}\cdot\dfrac{1}{\cos^3x}=\dfrac{1}{\sin x\cos^3x}\\\\R=\dfrac{\cos x}{\sin x}+2\cdot\dfrac{\sin x}{\cos x}+\dfrac{\sin^3x}{\cos^3x}\\\\=\dfrac{\cos x\cos^3x}{\sin x\cos^3x}+\dfrac{2\sin x\cos^2x}{\cos x\sin x\cos^2x}+\dfrac{\sin^3x\sin x}{\cos^3x\sin x}\\\\=\dfrac{\cos^4x+2\sin^2x\cos^2x+\sin^4x}{\sin x\cos^3x}\\\\=\dfrac{(\cos^2x)^2+2\sin^2x\cos^2x+(\sin^2x)^2}{\sin x\cos^3x}$

$=\dfrac{(\cos^2x+\sin^2x)^2}{\sin x\cos^3x}=\dfrac{1}{\sin x\cos^3x}=L\\\\Used:\\\tan(a)=\dfrac{\sin(a)}{\cos(a)}\\\cot(a)=\dfrac{\cos(a)}{\sin(a)}\\\sec(a)=\dfrac{1}{\cos(a)}\\\sin^2a+\cos^2a=1\\(a+b)^2=a^2+2ab+b^2$

8 0

3 years ago

Factorise the following quadratic expressions; 6a-18a21x -3x²1/16 - p²25v²-1​

Factorise the following quadratic expressions; 6a-18a21x -3x²1/16 - p²25v²-1