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All Summer in a Day"All Summer in a Day" is one of Ray Bradbury's most famous stories because it touches on the theme of cruelty toward those who are different. One of the reasons why Bradbury's science fiction is so...

Rip Van WinkleRip Van Winkle is a carefree man. He will play any sport or game with the children of the village. He will fish for hours on end, even without the faintest hope of catching anything. He will do...

The Catcher in the RyeNote: The literal meaning of something means that the reader is looking at exactly how this thing is being used in the plot. While the figurative meaning of something is looking at deeper meanings...

A Separate PeaceThere are several examples of foreshadowing in Chapter 2, most of them having to do with Gene’s eventual resentment of Finny. The first example occurs when Finny wears the pink shirt to class....

HistoryThere were three main things that people would have thought that the US would lose if some other country annexed the Philippines after the SpanishAmerican War. First, people thought that the US...

To Kill a MockingbirdBy the end of the book, Scout has learned to empathize with people. There are three main people Scout learns to empathize with in the book: her father, Mayella Ewell, and Boo Radley. By the end of...

Social SciencesTo answer this question, we need to know two things. We need to know the definitions for inflation, deflation, and disinflation and we need to know the formula for calculating the rate of...

The Last CrossingWhen you speak of chapters 2130, you are speaking of the latter parts of the novel The Last Crossing by Guy Vanderhaeghe. You are right in suggesting that there are important parts to the plot...

Julius CaesarWithout the manipulation of Cassius, Brutus would not have thought of murdering Caesar. And most likely there would not have been an assassination at all, because Cassius needed Brutus as a...

MathGraph the function:

MathGraph the function:

MathGraph the function:

MathGraph the function:

MathGraph the function:

Math`sin(arctan(x))` let `theta = arctan(x)` => `tan(theta)=x= x/1` by using the right triangle we get the hypotenuse is `= sqrt(1+x^2)` so `sin(theta) = x/(sqrt(1+x^2))`

Math`cos(sin^(1)(2x)) = cos(cos^(1)(sqrt(1(2x)^2)))` `=cos(cos^(1)(sqrt(14x^2)))` `=sqrt(14x^2)`

MathTaking 3x as oppsite side and 1 as the adjacent side,the hypotenuse will be equal to `sqrt(9x^2+1).` So `tan^1(3x)=sec^1((sqrt(9x^2+1))/1)` `sec(tan^(1)(3x))=sec(sec^(1)(sqrt(9x^2+1)))`...

Math`sin(cos^(1)x) = sin(sin^(1)(sqrt(1x^2)))` `= sqrt(1x^2)`

MathTaking (x1) as oppsite side and 1 as the hypotenuse,the adjacent side will be equal to `sqrt(1(x1)^2)=sqrt(2xx^2)` `sin^1(x1)=sec^1(1/(sqrt(2xx^2)))`...

MathTaking x as adjacent side and 3 as the hypotenuse,adjacent side will be equal to `sqrt(3^2x^2)=sqrt(9x^2)` `cos^1(x/3)=tan^1(sqrt(9x^2)/x)` `tan(cos^1(x/3))=tan(tan^1(sqrt(9x^2)/x))`...

MathTaking 1 as oppsite side and x as the adjacent side,the hypotenuse will be equal to `sqrt(1+x^2).` `tan^1(1/x)=cot^1(x/1)=cot^1x` `cot(tan^1(1/x))=cot(cot^1x)=x`

MathTaking x as oppsite side and `sqrt(2)` as the adjacent side,the hypotenuse will be equal to `sqrt(x^2+2).` `tan^1(x/sqrt(2))=csc^1(sqrt(x^2+2)/x)`...

Mathu = sin(v) cos(v) = sqrt(1(sin(v))^2) = sqrt(1u^2) But v = arcsin(u), therefore cos(arcsin(u)) = sqrt(1u^2). Next replace u with (xh)/r to get: cos(arcsin((xh)/r)) = sqrt(1(xh)^2/r^2).

MathGraph both of the functions. g(x): f(x):

MathGraph both function g(x) f(x)

MathGraph the function

MathGraph the function:

MathGraph the function:

MathGraph the function:

MathGraph the function:

MathGraph the function:

MathGraph the function:

Mathsin and arcsin cancel. Leaving the answer to be 0.3

MathTan and arctan cancel This leaves to the answer to be `45 `

Mathcos and arccos cancel. This leaves the answer to be `0.1 `

Mathsin and arcsin cancel. This leaves the answer to be `0.2 `

MathBy definition, `arcsin x = y` if `sin y = x` . Replacing x by sin y, yields: `arcsin x = arcsin (sin y) = y` Hence, replacing y by `3pi` , yields: `arcsin(sin 3pi) = 3pi` You need to evaluate...

Math`arccos(cos(7pi/2))` this could be thought of as 3 whole rounds `(6pi/2) ` plus` pi/2` as `pi/2` is between [1,1] we don't need to do anything. the arccos is the inverse of cos, so...

MathTaking 3 as oppsite side and 4 as the adjacent side,the hypotenuse will be equal to `sqrt(3^2+4^2)=5.` `tan^1(3/4)=sin^1(3/5)` `sin(tan^1(3/4))=sin(sin^1(3/5))=3/5.`

MathTaking 4 as oppsite side and 5 as the hypotenuse ,adjacent side will be equal to `sqrt(5^24^2)=3.` `sin^1(4/5)=sec^1(5/3)` `sec(sin^1(4/5))=sec(sec^1(5/3)) = 5/3.`

MathTaking 2 as oppsite side and 1 as the adjacent side,the hypotenuse will be equal to `sqrt(2^2+1^2)=sqrt(5).` `tan^1(2)=cos^1(1/sqrt(5))` `cos(tan^1(2))=cos(cos1(1/sqrt(5)))=1/sqrt(5).`

Math`sqrt(5)/5=1/sqrt(5)` Taking adjacent as 1 and hypotenuse as `sqrt(5)` ,then the opposite will be `sqrt((sqrt5)^21)=2.` `cos^1(1/sqrt5)=sin^1(2/sqrt5).`...

MathTaking opposite side as 5 and hypotenuse as 13 ,then the adjacent side will be sqrt(13^25^2)=12. sin^1(5/13)=cos^1(12/13). cos(sin^1(5/13))=cos(cos^1(12/13))=12/13=0.923

Math`csc(tan^1(5/12)) = csc(tan^1(5/12))` Taking 5 as opposite side and 12 as adjacent side ,then the hypotenuse will be `sqrt(12^2+5^2)=13.` `tan^1(5/12)=csc^1(13/5)`...

Math`tan^1(3/5) = tan^1(3/5)` Taking 3 as opposite side and 5 as adjacent side , then the hypotenuse will be `sqrt(5^2+3^2)=sqrt(34).` `tan^1(3/5)=sec^1(sqrt(34)/5)` `sec(tan^1(3/5)) =...

Math`tan(sin^1(3/4))=tan(sin^1(3/4))` `sin^1(3/4)= tan^1(3/sqrt(7))` `tan(sin^1(3/4))=tan(tan^1(3/sqrt7))=3/sqrt7=1.134`

MathWe know that `cos(arccos(2/3)) = 2/3` Therefore, using trig identities, => `sin = sqrt(1  cos^2) = sqrt(1  4/9)` Thus, => `sin = sqrt(5/9) = sqrt(5)/3`

Math`cotx=1/tanx` Given, => `cot(tan^1(5/8)) ` We know that `cotx=1/tanx` Therefore, => `1/(tan(arctan(5/8))) ` tan and arctan cancel => `1/(5/8) ` Simplify => `8/5 `

MathFirst, find the inner expression. Set `cosx=sqrt3/2 ` because we need to find what value of x yields `sqrt3/2 ` Use a unit circle to find this value => `x=(5pi)/6 ` Draw a triangle, Given the...

Math`sec(sin^1 (sqrt(2)/2))` `Let theta = sin^1 (sqrt(2)/2)` so we need to find `sec (theta)` => `sin(theta) = (sqrt(2)/2)` =>` sin(theta) = 1/sqrt(2)` we know `sin^2(theta) +cos^2(theta)...