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sattari [20]
3 years ago
9

Which values ​​of xwould make a polynomial equal to zero if the factors of the polynomial were (x + 4) and (x + 8)? O A. 4 and -

8 O B. 4and 8 O C. -4 and -8 O D. -4 and 8
Mathematics
1 answer:
tatiyna3 years ago
4 0

Answer:

-4 and -8

Step-by-step explanation:

Set each of the 2 factors = to 0 and solve for x:

x + 4 = 0 ↔ x = -4

x + 8 = 0 ↔ x = -8

The x-values -4 and -8 are the roots (zeros) of a polynomial whose factors are (x + 4) and (x + 8).

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Use the number line to determine the unknown addend in the given number sentence.
Gnesinka [82]

Answer:

C

Step-by-step explanation:

-4+(-5)=-9

-4=-9+5

-4=-4......correct

5 0
3 years ago
Read 2 more answers
Help plz answer quick
Inessa [10]

Answer:

around 1.67 inches

Step-by-step explanation:

We have to divide 1.25 inches by 0.75 inches to get the scale factor.

8 0
2 years ago
If y varies inversely as the square of x and y=4 when x=5, find y when x is 2
Len [333]

need to kind K

Y=k x 1/x^2 = k/x^2

Find K when y=4 & x=5

Y=k/x^2 =

4=k/5^2=

4=k/25

K=4*25 =100

When x = 2

Y=100/2^2

Y=100/4

Y=25


5 0
3 years ago
A box in a supply room contains 24 compact fluorescent lightbulbs, of which 8 are rated 13-watt, 9 are rated 18-watt, and 7 are
Marrrta [24]

Answer:

a) There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

b) There is a 8.65% probability that all three of the bulbs have the same rating.

c) There is a 12.45% probability that one bulb of each type is selected.

Step-by-step explanation:

There are 24 compact fluorescent lightbulbs in the box, of which:

8 are rated 13-watt

9 are rated 18-watt

7 are rated 23-watt

(a) What is the probability that exactly two of the selected bulbs are rated 23-watt?

There are 7 rated 23-watt among 23. There are no replacements(so the denominators in the multiplication decrease). Then can be chosen in different orders, so we have to permutate.

It is a permutation of 3(bulbs selected) with 2(23-watt) and 1(13 or 18 watt) repetitions. So

P = p^{3}_{2,1}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = \frac{3!}{2!1!}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 3*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 0.1764

There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

(b) What is the probability that all three of the bulbs have the same rating?

P = P_{1} + P_{2} + P_{3}

P_{1} is the probability that all three of them are 13-watt. So:

P_{1} = \frac{8}{24}*\frac{7}{23}*\frac{6}{22} = 0.0277

P_{2} is the probability that all three of them are 18-watt. So:

P_{2} = \frac{9}{24}*\frac{8}{23}*\frac{7}{22} = 0.0415

P_{3} is the probability that all three of them are 23-watt. So:

P_{3} = \frac{7}{24}*\frac{6}{23}*\frac{5}{22} = 0.0173

P = P_{1} + P_{2} + P_{3} = 0.0277 + 0.0415 + 0.0173 = 0.0865

There is a 8.65% probability that all three of the bulbs have the same rating.

(c) What is the probability that one bulb of each type is selected?

We have to permutate, permutation of 3(bulbs), with (1,1,1) repetitions(one for each type). So

P = p^{3}_{1,1,1}*\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 3**\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 0.1245

There is a 12.45% probability that one bulb of each type is selected.

3 0
3 years ago
Pls, I need help! 20 points
Montano1993 [528]

Answer:

5, 5, 10, 15

Step-by-step explanation:

He spent 1/4 of 20 which is 5 on food, 1/2 of 20 which is 10 on music, so if you add those you get 15. 20-15 is equal to 5.

8 0
3 years ago
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