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3 years ago

Solve n3 + 2n2 - 15n = 0 by factoring.

2 answers:

azamat3 years ago

5 0

N³ - 2n² - 15n = 0

"n"

n (n² + 2n - 15) = 0

n = 0

n² + 2n - 15 = 0

Δ = (2)² - 4(1)(-15)
Δ = 4 + 60 = 64

n' = (-2+8) / 2 = 6/2 =3
n'' = (-2-8) / 2 = -10/2 = -5

Solution:

S {-5 , 0 , 3 }

Naily [24]3 years ago

5 0

N^3+2n^2-15n=0

n(n^2+2n-15)=0

n(n^2+5n-3n-15)=0

n(n(n+5)-3(n+5))=0

n(n-3)(n+5)=0

n={-5, 0, 3}

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3 years ago

Using the image, determine the length of each arc. m RC= m CBR =

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Given:

Given that the measure of ∠CDR = 85°

We need to determine the measure of $\widehat{RC}$ and $\widehat{CBR}$

Measure of arc RC:

Since, we know that if a central angle is formed by two radii of the circle then the central angle is equal to the intercepted arc.

Thus, we have;

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Thus, the measure of $\widehat{RC}$ is 85°

Measure of arc CBR:

We know that 360° forms a full circle and to determine the measure of arc CBR, let us subtract the values 360 and 85.

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3 years ago

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faltersainse [42]

Answer:

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Given that Country Motorbikes Incorporated finds that it costs $400 to produce each motorbike, and that fixed costs are $1600 per day.

The price function is p(x) = 700 − 5x, where p is the price (in dollars) at which exactly x motorbikes will be sold.

If x units are produced and sold we have

Costs for x units = variable costs *x +Fixed costs $= 400x+1600$

Sales revenue = no of units sold * price = $x(700-5x) = 700x-5x^2$

Profit funciton = P(x) = Sales revenue - Total cost

= $700x-5x^2-400x-1600\\=300x-5x^2-1600$

To get maximum profit we use derivative test I derivative =0 and II derivative =negative

$P'(x) = 300-10x\\p"(x) = -10$

Producing 30 units will maximize the profit.

Max profit

=P(30) = 2900

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