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nadezda [96]
4 years ago
9

Jessica can do 60 jumping jacks in 2 minutes. Dale can do 150 jumping jacks in 5 mintues. Are these rates equivalent? Explain yo

ur reasoning.
Mathematics
1 answer:
Alinara [238K]4 years ago
3 0
Yes they are equivalent, because they both do 30 jumping jacks per minute. If you divide 60/2=30 & 150/5=30. Hope this helps!
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3 years ago
For the function given below, find a formula for the Riemann sum obtained by dividing the interval [0,5] into n equal subinterva
sergij07 [2.7K]

Given

we are given a function

f(x)=x^2+5

over the interval [0,5].

Required

we need to find formula for Riemann sum and calculate area under the curve over [0,5].

Explanation

If we divide interval [a,b] into n equal intervals, then each subinterval has width

\Delta x=\frac{b-a}{n}

and the endpoints are given by

a+k.\Delta x,\text{ for }0\leq k\leq n

For k=0 and k=n, we get

\begin{gathered} x_0=a+0(\frac{b-a}{n})=a \\ x_n=a+n(\frac{b-a}{n})=b \end{gathered}

Each rectangle has width and height as

\Delta x\text{ and }f(x_k)\text{ respectively.}

we sum the areas of all rectangles then take the limit n tends to infinity to get area under the curve:

Area=\lim_{n\to\infty}\sum_{k\mathop{=}1}^n\Delta x.f(x_k)

Here

f(x)=x^2+5\text{ over the interval \lbrack0,5\rbrack}\Delta x=\frac{5-0}{n}=\frac{5}{n}x_k=0+k.\Delta x=\frac{5k}{n}f(x_k)=f(\frac{5k}{n})=(\frac{5k}{n})^2+5=\frac{25k^2}{n^2}+5

Now Area=

\begin{gathered} \lim_{n\to\infty}\sum_{k\mathop{=}1}^n\Delta x.f(x_k)=\lim_{n\to\infty}\sum_{k\mathop{=}1}^n\frac{5}{n}(\frac{25k^2}{n^2}+5) \\ =\lim_{n\to\infty}\sum_{k\mathop{=}1}^n\frac{125k^2}{n^3}+\frac{25}{n} \\ =\lim_{n\to\infty}(\frac{125}{n^3}\sum_{k\mathop{=}1}^nk^2+\frac{25}{n}\sum_{k\mathop{=}1}^n1) \\ =\lim_{n\to\infty}(\frac{125}{n^3}.\frac{1}{6}n(n+1)(2n+1)+\frac{25}{n}n) \\ =\lim_{n\to\infty}(\frac{125(n+1)(2n+1)}{6n^2}+25) \\ =\lim_{n\to\infty}(\frac{125}{6}(1+\frac{1}{n})(2+\frac{1}{n})+25) \\ =\frac{125}{6}\times2+25=66.6 \end{gathered}

So the required area is 66.6 sq units.

3 0
1 year ago
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