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Vaselesa [24]
3 years ago
12

Write the first 4 terms of the sequence defined by the given rule f(n)=n2 -1

Mathematics
1 answer:
Softa [21]3 years ago
5 0

Answer:

<h2>0, 3, 8, 15</h2>

Step-by-step explanation:

Substitute n = 1, n = 2, n = 3 and n = 4 to the equation f(n) = n² - 1:

f(1) = 1² - 1 = 1 - 1 = 0

f(2) = 2² - 1 = 4 - 1 = 3

f(3) = 3² - 1 = 9 - 1 = 8

f(4) = 4² - 1 = 16 - 1 = 15

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borishaifa [10]

its pithagoreon so

c^2= a^2+ b^2

c^2= 32^2 +24^2

c^2= 1,024+576

c^2= 1600

c=\sqrt{1600\\= 40

5 0
3 years ago
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Vedmedyk [2.9K]
Huh come again? I’m not sure what you just said
6 0
3 years ago
What is the function? F(x)=2x^2-3x+1
olasank [31]

Answer:

To satisfy the hypotheses of the Mean Value Theorem a function must be continuous in the closed interval and differentiable in the open interval.

Step-by-step explanation:

As f(x)=2x3−3x+1 is a polynomial, it is continuous and has continuous derivatives of all orders for all real x, so it certainly satisfies the hypotheses of the theorem.

To find the value of c, calculate the derivative of f(x) and state the equality of the Mean Value Theorem:

dfdx=4x−3

f(b)−f(a)b−a=f'(c)

f(x)x=0=1

f(x)x=2=3

Hence:

3−12=4c−3

and c=1.

4 0
3 years ago
Read 2 more answers
Find q if p = -12 and the quotient p/q = -3.
s2008m [1.1K]

The quotient to p/q = -3 is 4

3 0
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 &#10;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
4 0
3 years ago
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