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Agata [3.3K]
3 years ago
13

The polynomial p(x) = 5x^3 – 9x^2 - 6x + 8 has a known factor of (x + 1).

Mathematics
2 answers:
alekssr [168]3 years ago
4 0

Answer:

(x+1)(5x-4)(x-2)

Step-by-step explanation:

We know that (x+1) is a factor, so we can use synthetic division to factor it out of 5x^3-9x^2-6x+8.

Once factored, we get (x+1)(5x^2-14x+8)

Now all we need to do is factor(5x^2-14x+8)

(5x^2-14x+8)=(5x-4)(x-2)

This gives us the answer (x+1)(5x-4)(x-2).

anzhelika [568]3 years ago
3 0

Answer:

(x+1)(5x-4)(x-2)

Step-by-step explanation:

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A local club is arranging a charter flight to Hawaii. The cost of the trip is ​$586 each for 80 ​passengers, with a refund of​ $
professor190 [17]

Answer:

a) The number of passengers that will maximize the revenue received from the flight is 99.

b) The maximum revenue is $48,609.

Step-by-step explanation:

We have to analyse two cases to build a piecewise function.

If there are 80 or less passengers, we have that:

The cost of the trip is $586 for each passenger. So

R(n) = 586n

If there are more than 80 passengers.

There is a refund of $5 per passenger for each passenger in excess of 80. So the cost for each passenger is

R(n) = (586 - 5(n-80))n = -5n^{2} +400n + 586n = -5n^{2} + 986n.

So we have the following piecewise function:

R(n) = \left \{ {{586n}, n\leq 80 \atop {-5n^{2} + 986n}, n > 80} \right

The maxium value of a quadratic function in the format of y(n) = an^{2} + bn + c happens at:

n_{v} = -\frac{b}{2a}

The maximum value is:

y(n_{v})

So:

(a) Find the number of passengers that will maximize the revenue received from the flight.

We have to see if n_{v} is higher than 80.

We have that, for n > 80, R(n) = -5n^{2} + 986n, so a = -5, b = 986

The number of passengers that will maximize the revenue received from the flight is:

n_{v} = -\frac{b}{2a} = -\frac{986}{2(-5)} = 98.6

Rounding up, the number of passengers that will maximize the revenue received from the flight is 99.

(b) Find the maximum revenue.

This is R(99).

R(n) = -5n^{2} + 986n

R(99) = -5*(99)^{2} + 986*(99) = 48609

The maximum revenue is $48,609.

8 0
3 years ago
Select all possible choices-
Roman55 [17]

I know B is on possible choice for this and maybe d almost certain

7 0
4 years ago
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antoniya [11.8K]

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bsdk badmash haram khor

5 0
3 years ago
Which system of equations does not have a real solution?
dolphi86 [110]

Answer:

It's A.

Step-by-step explanation:

Let's look at option A:

From the second equation y =  -10 - x. Substituting in the first equation:

-10 - x = x^2 + 3x - 5

x^2 + 4x + 5 = 0

Checking the discriminant b^2 - 4ac we get 16 - 4*1*5 = -4 so there are no real roots. (A negative discriminant means no real roots).

So A has no real solution.

B.

x^2 + 3x - 5 =  (20 - 4x)/5 = 4 - 0.8x

x^2 +3.8x - 9 = 0

b^2 - 4ac = (3.8)^2 - 4*1*-9 = 50.44 (positive) so there are real roots.

C.

x^2 + 3x - 5 =  -9 - x

x^2 + 4x + 4 = 0

b^2 - 4ac =  4^2 - 4*1*4 = 0  so there are real roots.

7 0
3 years ago
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Please help me! 20 points!
TEA [102]

Answer:

12 1/4 or (Decimal: 12.25)

Step-by-step explanation:

(3 1/2) (3 1/2)

= 7/2 (3 1/2)

= 7/2 (7/2)

= 49/4

= 12 1/4

   I hope I helped you!!

8 0
3 years ago
Read 2 more answers
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