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Blababa [14]
2 years ago
10

The solution set for 7q2 − 28 = 0 is { }. (Separate the solutions with a comma) NextReset

Mathematics
2 answers:
SVETLANKA909090 [29]2 years ago
6 0
7q^2 - 28 = 0

7q^2 = 28 
q^2 = 4
q = +/-2
 solution set is {-2,2}
exis [7]2 years ago
4 0

Answer:

{ 2,-2}

Step-by-step explanation:

In order to solve this set we just have to first clear the "q":

7q^{2} -28=0\\7q^{2} -28+28=+28\\7q^{2} =28\\\frac{7q^{2}}{7}  =\frac{28}{7} \\q^{2}=4\\\sqrt{q^{2}} =\sqrt{4} \\q= 2, -2

So as we know the solution for a square root is always a negative and positive number, so the solutions ser for 7q2-28=0 is 2 and -2

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2 years ago
Based on the polynomial remainder theorem, what is the value of the function when x = 3?
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Answer:

64

Step-by-step explanation:

Evaluate x^4 + 3 x^3 - 6 x^2 - 12 x - 8 where x = 3:

x^4 + 3 x^3 - 6 x^2 - 12 x - 8 = 3^4 + 3×3^3 - 6×3^2 - 12×3 - 8

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81 + 81 - 54 - 36 - 8 = (81 + 81) - (54 + 36 + 8):

(81 + 81) - (54 + 36 + 8)

| 8 | 1

+ | 8 | 1

1 | 6 | 2:

162 - (54 + 36 + 8)

| 1 |  

| 5 | 4

| 3 | 6

+ | | 8

| 9 | 8:

162 - 98

| | 15 |  

| 0 | 5 | 12

| 1 | 6 | 2

- | | 9 | 8

| 0 | 6 | 4:

Answer:  64

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