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Dvinal [7]
3 years ago
8

5.1(3 + 2.2x) > –14.25 – 6(1.7x + 4)?

Mathematics
1 answer:
MaRussiya [10]3 years ago
5 0

Answer:

x > -2.5

Step-by-step explanation:

1) 5.1 (3 + 2.2x) > -14.25 - 6 (1.7x + 4)

2) 15.3 + 11.22x > -14.25 - 10.2x - 24

3) 15.3 + 11.22x > -38.25 - 10.2x

4) 11.22x + 10.2x > -38.25 - 15.3

5) 21.42x > -53.55

6) x > -2.5

Alternate forms

x > -5/2x or x > -2 1/2

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The expression of integral as a limit of Riemann sums of given integral \int\limits^5_b {1} \, x/(2+x^{3}) dxis 4 \lim_{n \to \infty}∑n(n+4i)/2n^{3}+(n+4i)^{3} from i=1 to i=n.

Given an integral \int\limits^5_b {1} \, x/(2+x^{3}) dx.

We are required to express the integral as a limit of Riemann sums.

An integral basically assigns numbers to functions in a way that describes displacement, area, volume, and other concepts that arise by combining infinite data.

A Riemann sum is basically a certain kind of approximation of an integral by a finite sum.

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\int\limits^b_a {f(x)} \, dx=\lim_{n \to \infty}∑f(a+iΔx)Δx ,here Δx=(b-a)/n

\int\limits^5_1 {x/(2+x^{3}) } \, dx=f(x)=x/2+x^{3}

⇒Δx=(5-1)/n=4/n

f(a+iΔx)=f(1+4i/n)

f(1+4i/n)=[n^{2}(n+4i)]/2n^{3}+(n+4i)^{3}

\lim_{n \to \infty}∑f(a+iΔx)Δx=

\lim_{n \to \infty}∑n^{2}(n+4i)/2n^{3}+(n+4i)^{3}4/n

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Learn more about integral at brainly.com/question/27419605

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