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3 years ago

6

The Candle Company is having its semiannual sale. All items are 40 percent off. If the original price of a candle basket is $120

, what is the sale price?

2 answers:

Strike441 [17]3 years ago

6 0

Do 120 x .40 and you get 48 so you subtract 120-48 which = $72

sukhopar [10]3 years ago

4 0

All items are 40% off, so their new price is 60% of the old price.
60% = .6
120*.6= $72

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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

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

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⇒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.

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