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tino4ka555 [31]
3 years ago
9

Find the greatest integer value of $b$ for which the expression $\frac{9x^3+4x^2+11x+7}{x^2+bx+8}$ has a domain of all real numb

ers.
Mathematics
1 answer:
vovangra [49]3 years ago
3 0

Consider fraction \dfrac{9x^3+4x^2+11x+7}{x^2+bx+8}. The domain of this fraction is x^2+bx+8\neq 0.

Find the discriminant of the the equation x^2+bx+8=0:

D=b^2-4\cdot 1\cdot 8=b^2-32.

There are choices:

1. If D>0, there are two solutions of the equation;

2. D=0, there is only unique solution of the equation;

3. D<0, there are no solutions.

If x^2+bx+8\neq 0 you should consider case 3, then
b^2-32.

Therefore, the greatest integer is 5.

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The sum of 6 even integers is 130 what is the 2nd number in the sequence
castortr0y [4]

Answer:

The 2nd number in the sequence is 2\left(\frac{100}{12}\right)+2

Step-by-step explanation:

Let the six even numbers be 2x,2x+2,2x+4,2x+6,2x+8,2x+10

We are given that The sum of 6 even integers is 130.

So,2x+2x+2+2x+4+2x+6+2x+8+2x+10=130

12x+30=130

12x=100

x=\frac{100}{12}

So,the 2nd number in the sequence is 2\left(\frac{100}{12}\right)+2

6 0
4 years ago
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A bag contains 8 red marbles, 12 yellow marbles, 10 purple marbles, 9 green marblesand 22 orange marbles. You draw 5 marbles out
Allisa [31]

Answer:

13

Step-by-step explanation:

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4 0
3 years ago
PLEASE HELP ME
Serjik [45]
4 electrons will be on the second energy level. Only 2 can go on the first energy level, and only 4 can go on the second energy level. There will be 1 electron on the 3rd energy level. 
8 0
3 years ago
A stadium has 55,000 seats. Seats sell for $42 in Section A, $24 in Section B, and $18 in Section C. The number of seats in Sect
UkoKoshka [18]

Answer:

Section A = 27,500 seats

Section B = 14,800 seats

Section C = 12,700 seats

Step-by-step explanation:

Section A = $42

Section B = $24

Section C = $18

Revenue = $1,738,800

Total number of seats = 55,000

The number of seats in Section A equals the total number of seats in Sections B and C

A = B + C

A + B + C = 55,000

42A + 24B + 18C = $1,738,800

Substitute A = B + C into the equations

B + C + B + C = 55,000

42(B + C) + 24B + 18C = $1,738,800

2B + 2C = 55,000

42B + 42C + 24B + 18C = 1,738,800

2B + 2C = 55,000

66B + 60C = 1,738,800

Multiply (1) by 30

60B + 60C = 1,650,000. (1)

66B + 60C = 1,738,800. (2)

Subtract (1) from (2)

66B - 60B = 1,738,800 - 1,650,000

6B = 88,800

B = 88,800/6

= 14,800

B = 14,800

Substitute the value of B into

2B + 2C = 55,000

2(14,800) + 2C = 55,000

29,600 + 2C = 55,000

2C = 55,000 - 29,600

2C = 25,400

C = 25,400/2

= 12,700

C = 12,700

Substitute the values of B and 6 into

A + B + C = 55,000

A + 14,800 + 12,700 = 55,000

A + 27,500 = 55,000

A = 55,000 - 27,500

= 27,500

A = 27,500

Section A = 27,500 seats

Section B = 14,800 seats

Section C = 12,700 seats

3 0
3 years ago
Read 2 more answers
In math, how do you solve 2h&lt;200?
Svetlanka [38]

Answer:

h < 100

Step-by-step explanation:

isolate h by dividing both sides by 2

h < 100


8 0
3 years ago
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