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mamaluj [8]
2 years ago
12

For a sequence that starts with 5 and had a common difference of 2, what is the nth term

Mathematics
1 answer:
Charra [1.4K]2 years ago
3 0

Answer:

a_{n} = 2n + 3

Step-by-step explanation:

This is an arithmetic sequence with nth term

a_{n} = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Here a₁ = 5 and d = 2 , then

a_{n} = 5 + 2(n - 1) = 5 + 2n - 2 = 2n + 3

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H(t)=-16(t-1)^2+16 <br> What is the maximum height the ball reaches and how long does it take?
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Given t = time and h = height

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In the equation, h is 1 and k is 16.

So the maximum value is 16 at t = 1.

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2 years ago
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Identify whether the following equation has a unique solution, no solution, or infinitely many solutions.
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3 0
2 years ago
Consider the following division of polynomials.
Bond [772]

x^4=x^2\cdot x^2. Multiplying the denominator by x^2 gives

x^2(x^2+2x+8)=x^4+2x^3+8x^2

Subtracting this from the numerator gives a remainder of

(x^4+x^3+7x^2-6x+8)-(x^4+2x^3+8x^2)=-x^3-x^2-6x+8

-x^3=-x\cdot x^2. Multiplying the denominator by -x gives

-x(x^2+2x+8)=-x^3-2x^2-8x

and subtracting this from the previous remainder gives a new remainder of

(-x^3-x^2-6x+8)-(-x^3-2x^2-8x)=x^2+2x+8

This last remainder is exactly the same as the denominator, so x^2+2x+8 divides through it exactly and leaves us with 1.

What we showed here is that

\dfrac{x^4+x^3+7x^2-6x+8}{x^2+2x+8}=x^2-\dfrac{x^3+x^2+6x-8}{x^2+2x+8}

=x^2-x+\dfrac{x^2+2x+8}{x^2+2x+8}

=x^2-x+1

and this last expression is the quotient.

To verify this solution, we can simply multiply this by the original denominator:

(x^2+2x+8)(x^2-x+1)=x^2(x^2-x+1)+2x(x^2-x+1)+8(x^2-x+1)

=(x^4-x^3+x^2)+(2x^3-2x^2+2x)+(8x^2-8x+8)

=x^4+x^3+7x^2-6x+8

which matches the original numerator.

3 0
2 years ago
Read 2 more answers
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