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By calculus,

$[\mathcal{R}_1] = 30 + \frac{1}{4} \cdot \pi \cdot 10^2 + \int_0^5 \sqrt{100 - x^2} \ dx + \int_0^6 \sqrt{100 - x^2} \ dx.$

We can write out the areas similarly, to get [R_1] - [R_2] - [R_3] + [R_4] = 92.

it is not a perfect cube

I do not under stand the second part of your question....   divideng by .912847483    will make it a perfect cube of 22

    dividinb by  .07776   will make it a perfect cube of 50 

        etc etc   etc 

Perimeter of rectangle is 20   so width is 4    making   CD = 5

Trapezium area = 6 x 1/2 (3+5) = 24 cm²

Recall that by pairing up divisors of n, we can show that the product of the positive integer factors of n is $n^{x/2}$. We divide this formula by n to get the product of the proper positive integer factors of n, and we obtain $\frac{n^\frac{x}{2}}{n} = n^{\frac{x}{2}-1} = n^\frac{x-2}{2}$. Therefore, a=1, b=-2, and c=2, so a+b+c=1.

i am tot sure

I don't understand what you want us to find. Is it x and y, or the radius and center.

Here is a photo to show you the triangles I was talking about.

What is the sum of (8 x 102 ) + (6 x 10) + 2 and (5 x 103 ) + (2 x 102 ) + (8 x 10) + 9?

Hello Guest!

The sums of (8 x 102 ) + (6 x 10) + 2 and (5 x 103 ) + (2 x 102 ) + (8 x 10) + 9

are 878 and 808.

  !

There are formulas for this, but this is how I would do it:

$y=arctan(m),\quad\;\; m=cos(x)\\ tan(y)=m\\ \frac{dm}{dy}=sec^2(y)\qquad \frac{dm}{dx}=-sin(x)\\~\\ \frac{dy}{dx}=\frac{dy}{dm}\cdot \frac{dm}{dx}\\ \frac{dy}{dx}=\frac{1}{sec^2y}\cdot -sin(x)\\ \frac{dy}{dx}=\frac{-sin(x)}{sec^2(arctan(cos(x))}$

Again there are formulas but I used a series of right angled triangles to simplify the denominator. 

And I came up with  a denominator of      $1+ cos^2(x)$       

The problem with my method is that I will only guarantee if for angles in the first quadrant. 

Since it is squared it probably does not matter.

(that is assuming I made no stupid errors)

So now I have.

$\frac{dy}{dx}=\frac{-sin(x)}{1+cos^2(x)}\\ \text{Wolfram Alpha gives the asnwer as} \\ \frac{dy}{dx}=\frac{-2sin(x)}{3+cos(2x)}\\~\\ BUT\\ \text{These two answers are rearrangements of the same thing.}$

LaTex:

y=arctan(m),\quad\;\; m=cos(x)\\
tan(y)=m\\
\frac{dm}{dy}=sec^2(y)\qquad \frac{dm}{dx}=-sin(x)\\~\\
\frac{dy}{dx}=\frac{dy}{dm}\cdot \frac{dm}{dx}\\
\frac{dy}{dx}=\frac{1}{sec^2y}\cdot -sin(x)\\
\frac{dy}{dx}=\frac{-sin(x)}{sec^2(arctan(cos(x))}

Side  a = (1/2) / cos(15º)

[9 / (9 - 7)]  x  7 =31.5 hours to fill the Vat.

Hint,

Simplify the denominator first.

then multiply top and bottom by the conjugate of the denominator.

For all real numbers x except x=0 and x=1 the function f is defined by f(x/x-1)=1/x . 

Suppose 0<=t<=pi/2. Given that f(sec^(2)t)=(g(theta))^2 for some trigonometric function g, find g.
Enter your answer as one of the six options: sin, cos, tan, sec, csc, cot

$f(\frac{x}{x-1})=\frac{1}{x}\qquad where \;\;\;x\ne0\quad, x\ne1\\ Given\;\; 0<=t <= \frac{\pi}{2}\\ and\\ f(sec^2t)=(g(\theta))^2\qquad \text{Determine the function } g$

[The LaTex less than or equal to command is not working ... that is weird]

I have also found this question confusing:

But this is what I have

$sec^2t=\frac{x}{x-1}\\ xsec^2t-sec^2t=x\\ xsec^2t-x=sec^2t\\ x(sec^2t-1)=sec^2t\\ x=\frac{sec^2t}{sec^2t-1}\\ so\\ f(sec^2t)=\frac{sec^2t-1}{sec^2t}=1-cos^2t=sin^2t\\ so\\g(t)=sin(t)\\ g(\theta)=sin(\theta)$

LaTex

f(\frac{x}{x-1})=\frac{1}{x}\qquad where \;\;\;x\ne0\quad, x\ne1\\   

Given\;\; 0<=t <= \frac{\pi}{2}\\
and\\
f(sec^2t)=(g(\theta))^2\qquad \text{Determine the function  } g

sec^2t=\frac{x}{x-1}\\
xsec^2t-sec^2t=x\\
xsec^2t-x=sec^2t\\
x(sec^2t-1)=sec^2t\\
x=\frac{sec^2t}{sec^2t-1}\\
so\\
f(sec^2t)=\frac{sec^2t-1}{sec^2t}=1-cos^2t=sin^2t\\
so\\g(t)=sin(t)\\
g(\theta)=sin(\theta)

(1260 / 21) * 4 = 240 degrees

The second equation shows b = 5-3a       put this into the first equation for b

a^2   + (5-3a)^2 = 25    expand

a^2   + 9a^2 -30a + 25 = 25   

10a^2 -30a         =0

a(10a-30) = 0

a = 0     and   3          then b =   5   and   -4     corrrespondingly  

So for that 55+82=137
180= sum of triangle so
180-137=43 degrees

Take 1/2  of the 'x' coefficient , square it and add it to the expression to 'complete the square

[(1/2)(6/5) ]² =   36/100      or   9/25     for the blank

x = 20 bill                                value = 20x

6x = 5 bill                               value = 5 (6x) = 30x        values summed = 1450

20x + 30x = 1450

50x = 1450

x = 29                then 5 bill = 6x = 174 bills

Draw a Venn diagram ....you should be able to see the answer:

Angles of a triangle sum to 180 ^o

55^o + 82^o + ? = 180^o      Solve for ? 

5 = F * R^6,   

16.875 =F * R^9, solve for F, R

F =320 / 729 - This is the first term

R =3 / 2 - This is the common ratio

15th term =320 / 729 * (3/2)^14 =32805 / 256