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3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!!!

Mathematics

1 answer:

Mandarinka [93]3 years ago

7 0

Answer: B. (0, 6) and [-6, -4)

Step-by-step explanation:

The question is: when is the function negative?

function is also f(x) which is also "y".

So the question is asking: when is the y-value negative?

the line is below the x-axis between -6 and -4 and between 0 and 6.

--> [-6, -4) and (0, 6)

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I wi mark you brainless and give point of this gets completed

Paraphin [41]

Answer:

1. (x + 5)(x +10)

2. (x - 7)( x - 8)

3. (x - 4)( x - 7)

4. 2(7x−1)(x−9)

5. (2x+9)(x+6)

6. (x−1)(x−2)

7. (2x+3)(x+8)

8. (9x−2)(x−7)

9. 6(x−6)(x+7)

10. (9x+10)(x−4)

11. (3x+7)(x+3)

12. (x−9)(x−9)

13. 4(9x−10)(x−8)

14. 3(x+3)(x−9)

15. 2(x−1)(x+3)

16. 2(3x−1)(x+2)

17. 5(x−2)(x−2)

18. 2(x−1)(x−2)

Step-by-step explanation:

I'm so sorry it took this long that made my brain hurt it was worth it

Sorry it took so long..

hope I helped!!

5 0

3 years ago

Suppose we want to choose 5 colors, without replacement, from 8 distinct colors.1. If the order is relevant, how many can be don

Charra [1.4K]

(1) From the information given, if we want to choose 5 colors from 8 distinct colors and the order in which the selection is made is relevant, then what we have is a permutation.

The formula is given as;

$nP_r=\frac{n!}{(n-r)!}$

This formula means we need to select/arrange r items out of a total of n items and the anwer derived would be the total number of arrangements possible.

Therefore, we would have;

$\begin{gathered} nP_r\Rightarrow_8P_5 \\ _8P_5=\frac{8!}{(8-5)!}\Rightarrow\frac{8!}{3!} \\ _8P_5=\frac{8\times7\times6\times\ldots1}{3\times2\times1}\Rightarrow\frac{40320}{6} \\ _8P_5=6720 \end{gathered}$

Therefore, if the order is relevant, this selection can be done in 6,720 ways.

(2) If the order is NOT relevant, then what we need to calculate is a combination and the formula is;

$_nC_r=\frac{n!}{(n-r)!r!}$

The formula can now be applied as follows;

$\begin{gathered} _nC_r\Rightarrow_8C_5 \\ _8C_5=\frac{8!}{(8-5)!\times5!} \\ _8C_5=\frac{8!}{3!\times5!}\Rightarrow\frac{8\times7\times6\times\ldots1}{(3\times2\times1)\times(5\times4\times\ldots1)} \\ _8C_5=\frac{40320}{6\times120} \\ _8C_5=56 \end{gathered}$

If the order is not relevant, then the selection can be done in 56 ways.

3 0

1 year ago

Calculate limits x>-infinity -2x^5-3x+1

Lera25 [3.4K]

Given:

The limit problem is:

$\lim_{x\to -\infty}(-2x^5-3x+1)$

To find:

The value of the given limit problem.

Solution:

We have,

$\lim_{x\to -\infty}(-2x^5-3x+1)$

In the function $-2x^5-3x+1$ , the degree of the polynomial is 5, which is an odd number and the leading coefficient is -2, which is a negative number.