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4 years ago

1. (4x-3)(3x^2+8x-5)

1 answer:

6 0

Here, you have to foil the equation.

Distribute 4x to all sides, then distribute the -3 to all sides. Next, combine like terms. If you do this, your answer will be

$12x^3+23x^2-44x+15$

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This one might not look familiar yet, but it's very similar to the distance problems you've already seen.

Debora [2.8K]

Answer:

176.

Step-by-step explanation:

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3 years ago

What is 584 divided by 73

aev [14]

It is 8 because you divided it

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3 years ago

Read 2 more answers

A ship traveling east at 45 mph is 15 mi from a harbor when another ship leaves the harbor traveling east at 60 mph. How long do

Harman [31]

Answer:

1 hour

Step-by-step explanation:

I find these easiest to work by considering the initial difference in distance and the speed at with that gap is closing.

The gap is 15 miles, the distance the first ship is from harbor when the second ship starts.

The rate of closure is the difference in the speeds of the two ships:

60 mph -45 mph = 15 mph

Then the closure time is ...

time = distance/speed

time = (15 mi)/(15 mi/h) = 1 h

It will take the second ship 1 hour to catch up to the first ship.

7 0

2 years ago

Ln bisects klm into two congruent angles measuring (3x-4) and (4x-27) Find klm

blsea [12.9K]

The answer would be 135 (B)

4 0

3 years ago

The manufacturer of a CD player has found that the revenue R (in dollars) is Upper R (p )equals negative 5 p squared plus 1 com

AleksAgata [21]

Answer:

The maximum revenue is $1,20,125 that occurs when the unit price is $155.

Step-by-step explanation:

The revenue function is given as:

$R(p) = -5p^2 + 1550p$

where p is unit price in dollars.

First, we differentiate R(p) with respect to p, to get,

$\dfrac{d(R(p))}{dp} = \dfrac{d(-5p^2 + 1550p)}{dp} = -10p + 1550$

Equating the first derivative to zero, we get,

$\dfrac{d(R(p))}{dp} = 0\\\\-10p + 1550 = 0\\\\p = \dfrac{-1550}{-10} = 155$

Again differentiation R(p), with respect to p, we get,

$\dfrac{d^2(R(p))}{dp^2} = -10$

At p = 155

$\dfrac{d^2(R(p))}{dp^2} < 0$

Thus by double derivative test, maxima occurs at p = 155 for R(p).

Thus, maximum revenue occurs when p = $155.

Maximum revenue

$R(155) = -5(155)^2 + 1550(155) = 120125$

Thus, maximum revenue is $120125 that occurs when the unit price is $155.

6 0

3 years ago