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AveGali [126]
3 years ago
15

Factorise:2x^4+x^3-14x^2-19x-6

Mathematics
1 answer:
tester [92]3 years ago
6 0

Answer:

(x+1)(2x+1)(x+2)(x−3)

Step-by-step explanation:

Factor 2x^4+x^3−14x^2−19x^−6

2x^4+x^3−14x^2−19x^−6

=(x+1)(2x+1)(x+2)(x−3)

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Derive these identities using the addition or subtraction formulas for sine or cosine: sinacosb=(sin(a+b)+sin(a-b))/2
Sergeu [11.5K]

Answer:

The work is in the explanation.

Step-by-step explanation:

The sine addition identity is:

\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b).

The sine difference identity is:

\sin(a-b)=\sin(a)\cos(b)-\cos(a)\sin(a).

The cosine addition identity is:

\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b).

The cosine difference identity is:

\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b).

We need to find a way to put some or all of these together to get:

\sin(a)\cos(b)=\frac{\sin(a+b)+\sin(a-b)}{2}.

So I do notice on the right hand side the \sin(a+b) and the \sin(a-b).

Let's start there then.

There is a plus sign in between them so let's add those together:

\sin(a+b)+\sin(a-b)

=[\sin(a+b)]+[\sin(a-b)]

=[\sin(a)\cos(b)+\cos(a)\sin(b)]+[\sin(a)\cos(b)-\cos(a)\sin(b)]

There are two pairs of like terms. I will gather them together so you can see it more clearly:

=[\sin(a)\cos(b)+\sin(a)\cos(b)]+[\cos(a)\sin(b)-\cos(a)\sin(b)]

=2\sin(a)\cos(b)+0

=2\sin(a)\cos(b)

So this implies:

\sin(a+b)+\sin(a-b)=2\sin(a)\cos(b)

Divide both sides by 2:

\frac{\sin(a+b)+\sin(a-b)}{2}=\sin(a)\cos(b)

By the symmetric property we can write:

\sin(a)\cos(b)=\frac{\sin(a+b)+\sin(a-b)}{2}

3 0
3 years ago
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Answer: $384.13 ( rounded)

Original number: $384.125

Step-by-step explanation:

Find 9.75% of Paul's original stock, to find out how much it increased.

9.75*350/100=34.125

Add $34.125 to his original stock.

$384.125

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I need help I will give brainliest
Klio2033 [76]

Answer:

4.4 × 10 ^− 6

Step-by-step explanation:

Brainliest please :)?

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