Simplify the expression 5(n^2+n)-3n(2n^2+4n-2)

Mathematics

1 answer:

slega [8]3 years ago

8 0

first, lets distribute the equation. 5(n^2+n)-3n(2n^2+4n-2) would become

5n^2 +5n -6n^3 -12n^2 +6n

Now we add the like terms.

-6n^3 - 7n^2 +11n

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The Hu family goes out for lunch, and the price of the meal is $52. The sales tax on the meal is 6%, and the

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Answer:

62.92

Step-by-step explanation:

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2 years ago

Find the solution of the square root of the quantity of x plus 2 plus 4 equals 8, and determine if it is an extraneous solution

Iteru [2.4K]

Consider the equation:

$\sqrt{x+2}+4 = 8$

Subtracting '4' from both the sides of the equation, we get as

$\sqrt{x+2}+4-4= 8-4$

$\sqrt{x+2}= 4$

Squaring on both the sides of the equation, we get

$(\sqrt{x+2})^2 = (4)^2$

$x+2 = 16$

Subtracting '2' from both the sides of the equation, we get

$x+2-2=16-2$

x=14

Since, An extraneous solution is a solution that arises from the solving process that is not really a solution at all. But, in this equation x=14 is the solution of the given equation.

Hence, it is not an extraneous solution.

4 0

3 years ago

Read 2 more answers

16) Please help with question. WILL MARK BRAINLIEST + 10 POINTS.

Katyanochek1 [597]

We will use the sine and cosine of the sum of two angles, the sine and consine of $\frac{\pi}{2}$ , and the relation of the tangent with the sine and cosine:

$\sin (\alpha+\beta)=\sin \alpha\cdot\cos\beta + \cos\alpha\cdot\sin\beta

\cos(\alpha+\beta)=\cos\alpha\cdot\cos\beta-\sin\alpha\cdot\sin\beta$

$\sin\dfrac{\pi}{2}=1,\ \cos\dfrac{\pi}{2}=0$

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

If you use those identities, for $\alpha=x,\ \beta=\dfrac{\pi}{2}$ , you get:

$\sin\left(x+\dfrac{\pi}{2}\right) = \sin x\cdot\cos\dfrac{\pi}{2} + \cos x\cdot\sin\dfrac{\pi}{2} = \sin x\cdot0 + \cos x \cdot 1 = \cos x$

$\cos\left(x+\dfrac{\pi}{2}\right) = \cos x \cdot \cos\dfrac{\pi}{2} - \sin x\cdot\sin\dfrac{\pi}{2} = \cos x \cdot 0 - \sin x \cdot 1 = -\sin x$

Hence:

$\tan \left(x+\dfrac{\pi}{2}\right) = \dfrac{\sin\left(x+\dfrac{\pi}{2}\right)}{\cos\left(x+\dfrac{\pi}{2}\right)} = \dfrac{\cos x}{-\sin x} = -\cot x$