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4 years ago

Perform the indicated operation to write given polynomial in standard form: (x^3+2x^2 − 3x − 1) +(4 − x − x^3)

Mathematics

1 answer:

Rzqust [24]4 years ago

5 0

Answer:

$\left(x^3+2x^2-3x-1\right)+\left(4-x-x^3\right)=2x^2-4x+3$

Step-by-step explanation:

To write any polynomial in standard form, you look at the degree of each term. You then write each term in order of degree, from highest to lowest, left to write.

To add the polynomials you need to:

Remove parentheses

$\left(x^3+2x^2-3x-1\right)+\left(4-x-x^3\right)=x^3+2x^2-3x-1+4-x-x^3$

Group like terms

$\left(x^3+2x^2-3x-1\right)+\left(4-x-x^3\right)=x^3-x^3+2x^2-3x-x-1+4$

Add similar elements

$\left(x^3+2x^2-3x-1\right)+\left(4-x-x^3\right)=2x^2-3x-x-1+4\\\left(x^3+2x^2-3x-1\right)+\left(4-x-x^3\right)=2x^2-4x-1+4\\\left(x^3+2x^2-3x-1\right)+\left(4-x-x^3\right)=2x^2-4x+3$

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Five of Lucy's friends decided to split the cost of a $130 birthday gift. Then a sixth and seventh friend decided they also want

Effectus [21]

Answer:

$7.43

Step-by-step explanation:

First, solve for what each of the 5 friends are paying:

130/5 = 26

Then, solve for what each person is paying with the addition of 2 people:

130/7 = 18.57

Last, subtract to find the answer:

26-18.57 = 7.43

5 0

3 years ago

Find the discriminant and determine the number of real roots for each equation. Show steps.

Leni [432]

Answer:

discriminant: 4; number of real roots: 2

Step-by-step explanation:

The coefficients a, b and c of this quadratic are {-3, -2, 0}.

Thus, the discriminant b^2 - 4ac is (-2)^2 - 4(-3)(0), or 4.

Because the discriminant is positive, this quadratic has two real, unequal roots.

-3p^2 - 2p can be factored as follows: -p(3p + 2). If we set this expression equal to zero, we can find the roots: {-2/3, 0}

7 0

3 years ago

Which quadratic function is shown in the graph?<br> A) f(x) = x2

ANTONII [103]

Answer: a I think

Step-by-step explanation:

5 0

3 years ago

What is the estimate of 45.15

erastova [34]

Answer:

The estimate would be:

(one of them :)

45 (if your doing the entire number)

0 (if your rounding out of 100)

45.2 (nearest hundredth)

45.0 (nearest tenth)

50 (nearest tens)

Hopefully one of these works for you.

6 0

3 years ago

Ah! Okay, need help solving 14, and just checking for 16 and 4.

Allisa [31]

4. Correct. You also could have used the limit test for divergence for the same conclusion (the summand approaches infinity).

- - -

14. I'm guessing the instructions are the same as for 16. Rewrite as

$f(x)=\dfrac4{2x+3}=\dfrac{\frac43}{1-\left(-\frac{2x}3\right)}$

Now recall that for $|x|$ , we have

$\dfrac1{1-x}=\displaystyle\sum_{n\ge0}x^n$

so that for this function, we get

$f(x)=\dfrac43\displaystyle\sum_{n\ge0}\left(-\frac{2x}3\right)^n$

Because this is a geometric sum, this converges when $\left|-\dfrac{2x}3\right|$ , or $|x|$ . This would be the interval of convergence.

Your hunch about checking the endpoints is correct. Checking is easy in this case, because at the endpoints (-3/2 and 3/2) the series obviously diverges.

- - -

16. This one is kind of tricky, and there's more than one way to do it. The standard method would be to take the antiderivative:

$F(x)=\displaystyle\int f(x)\,\mathrm dx=\int\frac{\mathrm dx}{(1+x)^2}=-\frac1{1+x}+C$

We also have

$\displaystyle-\frac1{1+x}=-\frac1{1-(-x)}=-\sum_{n\ge0}(-x)^n\implies F(x)=C-1-\sum_{n\ge1}(-x)^n$

and differentiating this gives

$f(x)=-\displaystyle\sum_{n\ge1}n(-x)^{n-1}=-\sum_{n\ge0}(n+1)(-x)^n=\sum_{n\ge0}(n+1)(-1)^{n+1}x^n$

By the ratio test, this converges when

$\displaystyle\lim_{n\to\infty}\left|\frac{(n+2)(-1)^{n+2}x^{n+1}}{(n+1)(-1)^{n+1}x^n}\right|$

The limit reduces to

$\displaystyle|x|\lim_{n\to\infty}\frac{n+2}{n+1}=|x|$

and so the series converges absolutely for $|x|$ . Checking the endpoints is also easy in this case. The factor of $n+1$ is a clear sign that the series will diverge at either extreme.

7 0

3 years ago