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ratelena [41]
3 years ago
7

How do I divide 2x^4 + 5x^3 + x - 1 by x^2 - 2x + 1 using long division?

Mathematics
1 answer:
Elza [17]3 years ago
7 0

Answer: 2x^2+ 9x +16+ 24x- 17/ x^2- 2x +1

Step-by-step explanation:

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3 years ago
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Klio2033 [76]

Answer:

51%

Step-by-step explanation:

Add up the tallies, as follows:  17+24+9+19.  Then divide the "cycling" tally (24) by this sum ( 69), obtaining the "observed probability that the customer will want to cycle"):  p = 24/69 = 34.7%.

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3 years ago
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Nookie1986 [14]

Step-by-step explanation:

\lim_{n \to \infty} \sum\limits_{k=1}^{n}f(x_{k}) \Delta x = \int\limits^a_b {f(x)} \, dx \\where\ \Delta x = \frac{b-a}{n} \ and\ x_{k}=a+\Delta x \times k

In this case we have:

Δx = 3/n

b − a = 3

a = 1

b = 4

So the integral is:

∫₁⁴ √x dx

To evaluate the integral, we write the radical as an exponent.

∫₁⁴ x^½ dx

= ⅔ x^³/₂ + C |₁⁴

= (⅔ 4^³/₂ + C) − (⅔ 1^³/₂ + C)

= ⅔ (8) + C − ⅔ − C

= 14/3

If ∫₁⁴ f(x) dx = e⁴ − e, then:

∫₁⁴ (2f(x) − 1) dx

= 2 ∫₁⁴ f(x) dx − ∫₁⁴ dx

= 2 (e⁴ − e) − (x + C) |₁⁴

= 2e⁴ − 2e − 3

∫ sec²(x/k) dx

k ∫ 1/k sec²(x/k) dx

k tan(x/k) + C

Evaluating between x=0 and x=π/2:

k tan(π/(2k)) + C − (k tan(0) + C)

k tan(π/(2k))

Setting this equal to k:

k tan(π/(2k)) = k

tan(π/(2k)) = 1

π/(2k) = π/4

1/(2k) = 1/4

2k = 4

k = 2

8 0
3 years ago
Help please!! 15 points!!
maxonik [38]

Answer:

A and D

Step-by-step explanation:

One side-base right triangle and one angle-based right triangle.

6 0
2 years ago
Read 2 more answers
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