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Natali [406]
2 years ago
12

Which of the following could represent the graph of f(x) = y3 – x2 - 6x?

Mathematics
1 answer:
professor190 [17]2 years ago
6 0

Answer: A

Step-by-step explanation:

x^{3}-x^{2}-6x=0 \\ \\ x(x^{2}-x-6)=0 \\ \\ x(x-3)(x+2)=0 \\ \\ x=-2, 0, 3

So there needs to be roots at x=-2, 0, and 3.

Also, the end behavior needs to approach infinity as x approaches infinity and negative infinity as x approaches negative infinity (since the leading coefficient is positive). This rules out the last option.

Now, we can also use the fact that the maximum number of turning points is one less than the degree, for a maximum of 2 turning points. This rules out the middle option as well.

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Select the two values of x that are roots of this equation. 2x2 + 7x + 6 = 0
Anna007 [38]

Answer:

x = -3/2, -2

Step-by-step explanation:

2x^2 + 7x + 6 = 0

(2x + 3)(x + 2) = 0

(2x + 3)

2x = -3

x = -3/2

(x + 2)

x = -2

7 0
2 years ago
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How many whole-number solutions does the inequality |x-5|≤8 have?
-BARSIC- [3]
The answer is -3 to 13
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3 years ago
A square with a side length of 5 inches has a area of 25 in<br><br> true or false
Goryan [66]

Answer:

TRUE

Step-by-step explanation:

5 x 5 = 25 in²

6 0
3 years ago
Can I get help with finding the Fourier cosine series of F(x) = x - x^2
trapecia [35]
Assuming you want the cosine series expansion over an arbitrary symmetric interval [-L,L], L\neq0, the cosine series is given by

f_C(x)=\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos nx

You have

a_0=\displaystyle\frac1L\int_{-L}^Lf(x)\,\mathrm dx
a_0=\dfrac1L\left(\dfrac{x^2}2-\dfrac{x^3}3\right)\bigg|_{x=-L}^{x=L}
a_0=\dfrac1L\left(\left(\dfrac{L^2}2-\dfrac{L^3}3\right)-\left(\dfrac{(-L)^2}2-\dfrac{(-L)^3}3\right)\right)
a_0=-\dfrac{2L^2}3

a_n=\displaystyle\frac1L\int_{-L}^Lf(x)\cos nx\,\mathrm dx

Two successive rounds of integration by parts (I leave the details to you) gives an antiderivative of

\displaystyle\int(x-x^2)\cos nx\,\mathrm dx=\frac{(1-2x)\cos nx}{n^2}-\dfrac{(2+n^2x-n^2x^2)\sin nx}{n^3}

and so

a_n=-\dfrac{4L\cos nL}{n^2}+\dfrac{(4-2n^2L^2)\sin nL}{n^3}

So the cosine series for f(x) periodic over an interval [-L,L] is

f_C(x)=-\dfrac{L^2}3+\displaystyle\sum_{n\ge1}\left(-\dfrac{4L\cos nL}{n^2L}+\dfrac{(4-2n^2L^2)\sin nL}{n^3L}\right)\cos nx
4 0
3 years ago
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dalvyx [7]

Answer:

AE

Step-by-step explanation:

pls make me brainliest

3 0
3 years ago
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