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3 years ago

Please can you help me I'll mark brainiest please and it's being timed!!

Mathematics

1 answer:

Scrat [10]3 years ago

8 0

Answer:

Step-by-step explanation:

The point ( -2 , 3) lies in the 2nd quadrant

Point are generally in the form (x , y). The point you need to plot is ( -2 , 3) which means the value of x = -2 and the value of y = 3.

I have attached an image you can check it out.

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3 years ago

Sixteen times a number j is no less than -2

labwork [276]

16x2=32 hope This helps

8 0

3 years ago

Convert 11pi/15 to degrees

Leto [7]

Answer:

-132

Step-by-step explanation:

To convert radians to degrees, multiply by 180π , since a full circle is 360° or 2π radians. Cancel the common factor of π . Move the leading negative in −11π15 - 11 π 15 into the numerator.

3 0

4 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.

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1 year ago

1. Write the point-slope form of the equation of the line with a slope of 2 that

katrin2010 [14]

Answer:

y+4=2(x-4)

Step-by-step explanation:

y-y1=m(x-x1)

y-(-4)=2(x-4)

y+4=2(x-4) point slope form

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3 years ago