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4 years ago

Which statement about this figure is true?

Mathematics

1 answer:

fgiga [73]4 years ago

5 0

**IT HAS NO ROTATIONAL SYMMETRY**

-ON GRADPOINT

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4 years ago

Can someone help me with my last homework math related problem thank you.

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3 years ago

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Which example supports the given conjecture?

Diano4ka-milaya [45]

The example A. 3+5+7=15 is the write answer

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4 years ago

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Given the following equation x2 + 4x = 21 find the solutions. (Show All Work)

NeTakaya

Step-by-step explanation:

15a)

$x^2+4x=21$ Subtract both sides by 21

$x^2+4x-21 =0$

15b)

7 and -3 multiply to -21 and add to 4

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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