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2 years ago

3) Which of the equivalent expressions for P(x) reveals the profit when the price is

Mathematics

1 answer:

kondor19780726 [428]2 years ago

6 0

Question:

Which of the equivalent expressions for P(x) reveals the profit when the price is

zero without changing the form of the expression? P(x) = -3x² + 432x - 1680

a) P(x) = -3(x - 4)(x - 140)

b) P(x) = -3(x - 72)² + 13872

Find the profit when the price is zero

Answer:

- P(x) = -3(x - 140)(x - 4).

- P(0) = -1680

Step-by-step explanation:

Given

P(x) = -3x² + 432x - 1680

Required

- Find equivalent of P(x)

- Find P(x) when x = 0

To find the equivalent of P(x), we simply factorize P(x) = -3x² + 432x - 1680.

This is done as follows

P(x) = -3x² + 432x - 1680

P(x) = -3x² + 12x + 420x - 1680

P(x) = -3x(x - 4) + 420(x - 4)

P(x) = (-3x + 420)(x - 4)

Further expand -3x + 420

P(x) = -3(x - 140)(x - 4).

Hence, the equivalent of

P(x) = -3x² + 432x - 1680

without changing its form is

P(x) = -3(x - 140)(x - 4).

When price is 0, then x = 0.

Substitute 0 for P(x) in P(x) = -3(x - 140)(x - 4).

P(0) = -3(0 - 140)(0 - 4).

P(0) = -3(-140)(-4)

P(0) = -1680

Hence, the profit when price is 0 is -1680

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2 years ago

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MA_775_DIABLO [31]

There's a slight problem with your question, but we'll get to that...

Consecutive terms of the sequence are separated by a fixed difference of 21 (22 = 1 + 21, 43 = 22 + 21, 64 = 43 + 21, and so on), so the n-th term of the sequence, a (n), is given recursively by

• a (1) = 1

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We can find the explicit rule for the sequence by iterative substitution:

a (2) = a (1) + 21

a (3) = a (2) + 21 = (a (1) + 21) + 21 = a (1) + 2×21

a (4) = a (3) + 21 = (a (1) + 2×21) + 21 = a (1) + 3×21

and so on, with the general pattern

a (n) = a (1) + 21 (n - 1) = 21n - 20

Now, we're told that the sum of some number N of terms in this sequence is 2332. In other words, the N-th partial sum of the sequence is

a (1) + a (2) + a (3) + … + a (N - 1) + a (N) = 2332

or more compactly,

$\displaystyle\sum_{n=1}^N a(n) = 2332$

It's important to note that N must be some positive integer.

Replace a (n) by the explicit rule:

$\displaystyle\sum_{n=1}^N (21n-20) = 2332$

Expand the sum on the left as

$\displaystyle 21 \sum_{n=1}^N n-20\sum_{n=1}^N1 = 2332$

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$\displaystyle\sum_{k=1}^n1=\underbrace{1+1+\cdots+1}_{n\text{ times}}=n$

$\displaystyle\sum_{k=1}^nk=1+2+3+\cdots+n=\frac{n(n+1)}2$

So the sum of the first N terms of a (n) is such that

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N = (19 ± √392,137)/42 ≈ -14.45 or 15.36

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