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3 years ago

Fill in the blank with the appropriate word(s)

Mathematics

1 answer:

Naddik [55]3 years ago

8 0

Answer:

Step-by-step explanation:

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3x3-27x=0<br> Solve by factoring a

Eddi Din [679]

After factoring $3x^3-27x=0$ the solutions are x=0 , x=3, x= -3

Step-by-step explanation:

We need to solve by factoring:

$3x^3-27x=0$

Solving:

$3x^3-27x=0$

Taking 3x common:

$3x(x^2-9)=0$

Now using formula $a^2-b^2=(a-b)(a+b)$

$3x((x)^2-(3)^2)=0\\3x(x-3)(x+3)=0$

$3x=0\,\,and\,\,x+3=0\,\,and\,\,x-3=0\\x=0\,\,and\,\,x=-3\,\,and\,\,x=3\\$

So, After factoring $3x^3-27x=0$ the solutions are x=0 , x=3, x= -3

Keywords: Factorization

Learn more about Factorization at:

#learnwithBrainly

5 0

3 years ago

E help me pls pls plzzzzzzzz

LuckyWell [14K]

Answer:

What is part 1?

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Use the power reduction formulas to rewrite the expression. (Hint: Your answer should not contain any exponents greater than 1.)

Lapatulllka [165]

Some useful relations and identities:

$\tan x=\dfrac{\sin x}{\cos x}$

$\sin^2x=\dfrac{1-\cos2x}2$

$\cos^2x=\dfrac{1+\cos2x}2$

By the first relation, we have

$\tan^2x\sin^3x=\dfrac{\sin^5x}{\cos^2x}=\dfrac{(\sin^2x)^2\sin x}{\cos^2x}$

Applying the two latter identities, we get

$\dfrac{\left(\frac{1-\cos2x}2\right)^2\sin x}{\frac{1+\cos2x}2}=\dfrac{\frac{1-2\cos2x+\cos^22x}4\sin x}{\frac{1+\cos2x}2}=\dfrac{(1-2\cos2x+\cos^22x)\sin x}{2(1+\cos2x)}$

We can apply the third identity again:

$\dfrac{(1-2\cos2x+\cos^22x)\sin x}{2(1+\cos2x)}=\dfrac{\left(1-2\cos2x+\frac{1+\cos4x}2\right)\sin x}{2(1+\cos2x)}=\dfrac{(3-4\cos2x+\cos4x)\sin x}{4(1+\cos2x)}$