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3 years ago

What whole numbers lie between 39

Mathematics

2 answers:

ArbitrLikvidat [17]3 years ago

8 0

Answer
6 and 7

Explanation

Ludmilka [50]3 years ago

6 0

Answer: none

Step-by-step explanation: there are no WHOLE numbers between 39

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Express the integral as a limit of Riemann sums. Do not evaluate the limit. (Use the right endpoints of each subinterval as your

Veronika [31]

The expression of integral as a limit of Riemann sums of given integral $\int\limits^5_b {1} \, x/(2+x^{3}) dx$ is 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.

Using Riemann sums, we have :

$\int\limits^b_a {f(x)} \, dx$ = $\lim_{n \to \infty}$ ∑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=

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

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

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

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

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2 years ago

Which answer best describes the complex zeros of the polynomial function?

marysya [2.9K]

Answer-

D. The function has one real zero and two nonreal zeros. The graph of the function intersects the x-axis at exactly one location.

Solution-

The given polynomial is,

$f(x)=x^3-x^2+6x-6$

The zeros of the polynomials are,

$\Rightarrow f(x)=0$

$\Rightarrow x^3-x^2+6x-6=0$

$\Rightarrow x^2(x-1)+6(x-1)=0$

$\Rightarrow (x^2+6)(x-1)=0$

$\Rightarrow (x^2+6)=0,(x-1)=0$

$\Rightarrow x=\sqrt{-6},\ x=1$

$\Rightarrow x=1,\ \pm \sqrt6i$

Therefore, this function has only one real zero i.e 1 and two nonreal zeros i.e ±√6i . The graph of the function intersects the x-axis at exactly one location i.e at x = 1