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3 years ago

Given f(x) = 2x³ 3x and g(x) = -3x2 +1, what is the value of h(3) where h(x) = (f+g)(x)?

Mathematics

1 answer:

Anni [7]3 years ago

4 0

Answer:

Step-by-step explanation:

Operations of functions:

h(x) = f(z) + g(z)

h(x) = (2z³ - 3z) + (-3z² + 1)

Simplify or expand:

(2z³ - 3z) + (-3z² + 1)

Determine the sign

2z³ - 3z - 3z² + 1

Rewrite the expression

2z³ - 3z² - 3z + 1

Calculate:

Substitute z = 3 into

2z³ - 3z² - 3z + 1

Substitute

2 × 3³ - 3 × 3² - 3 × 3 + 1

Calculate

HOPE THIS HELPS AND HAVE A NICE DAY <3

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If A is a subset and equal to B then B'-A' is equal to

german

Answer:

A-B

Step-by-step explanation:

It doesn't say A=B... But just that A is a subset of B.

Anyways, why not try an example to help reveal the answer.

Let the universe set, U, be U={1,2,3,4,5,6}.

Let B={1,2,3,4} and A={1,3,5}

First choice would become A'={2,4,6}

Second choice would become B'={5,6}

Third choice would become A-B={5}

Fourth choice is just { }

B'-A'={5}

The only choice matching is A-B

*Just so you know X-Y is the set of elements contained in X only if they are not also in Y.

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3 years ago

A building is 50 feet tall. Imagine you are standing 30 feet away from it. What is the angle when you are looking up at your fri

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Answer:

53.1

Step-by-step explanation:

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4 years ago

I need a little help with this problem: 5k + 1 = 4k - 1

Varvara68 [4.7K]

5k-4k=-1+-1
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3 years ago

Evaluate Dx / ^ 9-8x - x2^

Solnce55 [7]

It depends on what you mean by the delimiting carats "^"...

Since you use parentheses appropriately in the answer choices, I'm going to go out on a limb here and assume something like "^x^" stands for $\sqrt x$ .

In that case, you want to find the antiderivative,

$\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}$

Complete the square in the denominator:

$9-8x-x^2=25-(16+8x+x^2)=5^2-(x+4)^2$

Now substitute $x+4=5\sin y$ , so that $\mathrm dx=5\cos y\,\mathrm dy$ . Then

$\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\int\frac{5\cos y}{\sqrt{5^2-(5\sin y)^2}}\,\mathrm dy$

which simplifies to

$\displaystyle\int\frac{5\cos 
y}{5\sqrt{1-\sin^2y}}\,\mathrm dy=\int\frac{\cos y}{\sqrt{\cos^2y}}\,\mathrm dy$

Now, recall that $\sqrt{x^2}=|x|$ . But we want the substitution we made to be reversible, so that

$x+4=5\sin y\iff y=\sin^{-1}\left(\dfrac{x+4}5\right)$

which implies that $-\dfrac\pi2\le y\le\dfrac\pi2$ . (This is the range of the inverse sine function.)

Under these conditions, we have $\cos y\ge0$ , which lets us reduce $\sqrt{\cos^2y}=|\cos y|=\cos y$ . Finally,

$\displaystyle\int\frac{\cos y}{\cos y}\,\mathrm dy=\int\mathrm dy=y+C$

and back-substituting to get this in terms of $x$ yields

$\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\sin^{-1}\left(\frac{x+4}5\right)+C$

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3 years ago

Please help if possible thank you

Helga [31]

$y = \frac{3}{2}$

5 0

3 years ago