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olasank [31]
3 years ago
11

For the function​ below, find a formula for the upper sum obtained by dividing the interval [a comma b ][a,b] into n equal subin

tervals. Then take a limit of these sums as n right arrow infinityn → [infinity] to calculate the area under the curve over [a comma b ][a,b]. f (x )equals x squared plus 3f(x)=x2+3 over the interval [0 comma 4 ]

Mathematics
1 answer:
Vlad [161]3 years ago
8 0

Answer:

See below

Step-by-step explanation:

We start by dividing the interval [0,4] into n sub-intervals of length 4/n

[0,\displaystyle\frac{4}{n}],[\displaystyle\frac{4}{n},\displaystyle\frac{2*4}{n}],[\displaystyle\frac{2*4}{n},\displaystyle\frac{3*4}{n}],...,[\displaystyle\frac{(n-1)*4}{n},4]

Since f is increasing in the interval [0,4], the upper sum is obtained by evaluating f at the right end of each sub-interval multiplied by 4/n.

Geometrically, these are the areas of the rectangles whose height is f evaluated at the right end of the interval and base 4/n (see picture)

\displaystyle\frac{4}{n}f(\displaystyle\frac{1*4}{n})+\displaystyle\frac{4}{n}f(\displaystyle\frac{2*4}{n})+...+\displaystyle\frac{4}{n}f(\displaystyle\frac{n*4}{n})=\\\\=\displaystyle\frac{4}{n}((\displaystyle\frac{1*4}{n})^2+3+(\displaystyle\frac{2*4}{n})^2+3+...+(\displaystyle\frac{n*4}{n})^2+3)=\\\\\displaystyle\frac{4}{n}((1^2+2^2+...+n^2)\displaystyle\frac{4^2}{n^2}+3n)=\\\\\displaystyle\frac{4^3}{n^3}(1^2+2^2+...+n^2)+12

but  

1^2+2^2+...+n^2=\displaystyle\frac{n(n+1)(2n+1)}{6}

so the upper sum equals

\displaystyle\frac{4^3}{n^3}(1^2+2^2+...+n^2)+12=\displaystyle\frac{4^3}{n^3}\displaystyle\frac{n(n+1)(2n+1)}{6}+12=\\\\\displaystyle\frac{4^3}{6}(2+\displaystyle\frac{3}{n}+\displaystyle\frac{1}{n^2})+12

When n\rightarrow \infty both \displaystyle\frac{3}{n} and \displaystyle\frac{1}{n^2} tend to zero and the upper sum tends to

\displaystyle\frac{4^3}{3}+12=\displaystyle\frac{100}{3}

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Dilation is the process in which the dimensions of a given shape is increased or decreased by a scale factor. Therefore the answers to the questions are:

A. Since the dilation of triangle XYZ do not affect the measure of its internal angles, then the trigonometric ratios of respective internal angles are the same.

B. segment CB = 10

ii. segment AB = 11.18

From the given information in the question, triangle XYZ is an image of triangle ACB. In triangle XYZ, given that Sin X = ; it implies that its hypotenuse is 5.59, and the opposite side as 5. So that;

Sin X =

X =  0.8945

X  =

Thus,

<X + <Y + <Z =

+  + <Z =

<Z =  -

   =

<Z =

So then, the third side can be determined by applying the Pythagoras theorem.

=  +

=  +

=  -

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x =

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x = 2.5

Thus,

Cos X =

          =

Cos X =

Tan X =

         =

Tan X =

Therefore:

A. Considering triangles XYZ and ACB, the measure of their respective internal angles are the same.

i.e <X ≅ <A

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   <C ≅ <Y

So that the trigonometric ratios of respective internal angles are the same.

B. Given that triangle XYZ was dilated by a scale factor of 2, then the respective length of each sides of triangle ACB is twice that of XYZ.

Then,

i. segment CB = 2 x segment YZ

                    = 2 x 5

segment CB = 10

ii. segment AB = 2 x segment XZ

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8 0
2 years ago
PLEASEE HELP!!<br><br> What is the value of x?<br><br> Enter your answer in the box
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First we need to figure out how long YK is and the best way to do this is to set up a proportion

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then add the two parts of the line

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