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the best answer is the one with the most likes

Here is a different approach.The arc AC has measure \(\frac{1}{3}\cdot2\pi\) and the angle AOB has measure half of the arc or \(\frac {\pi}{3}\). If the radius of the larger circle is \(R \), then the measure of OB, the radius of the smaller circle, is \(Rcos(\frac{\pi}{3})\)and so the orange region has area \(\pi(Rcos(\frac{\pi}{3}))^2=\frac{1}{4}\pi R^2\) . The blue region, however, has area equal to \(\frac{1}{3}\)of the difference between area of the larger circle and the area of the smaller circle, i.e. \(\frac{1}{3}(\pi R^2-\frac{1}{4}\pi R^2) =\frac{1}{4}\pi R^2\). So the two regions have the same area, each being \(\frac{1}{4}\) of the area of the larger circle.

its ok

Calculate the hypotenuse of the right triangle formed by negative x axis and the point and the origin

hyp^2 = (-9)^2 + 10^2 = 181

hyp = sqrt (181)

Sin = opp/hyp      (use reference angle  that is negative x axis and the hypotenuse)

      10/sqrt181     or 10 sqrt181 / 181

Second one

The vertex will be at   ([4 + 2] / 2, 6 )  =   ( 3, 6) =  (h, k)

This parabola opens to the left   because the directrix is to the right of the focus

The form is

-4a ( x - h)  = (y - k)^2

a = the distance between the vertex and focus  =  1

So  we have that

-4 (1) (x - 3)  = (y - 6)^2

-4 (x - 3)  = (y - 6)^2

Here's the graph  :  https://www.desmos.com/calculator/9wy9pmdttx

3x^2 + 24x  - 2y + 52  =  0       rearrange as

2y  =  3x^2 + 24x  + 52           divide both sides by  2

y = (3/2)x^2 + 12x + 26          factor out the (3/2)

y  =  (3/2)  [ x^2  + 8x + 52/3 ]        

Complete the square on x........take 1/2  of 8  = 4  ......square it   =  16.....add and subtract within the parentheses

y = (3/2) [ x^2 + 8x + 16 + 52/3  - 16 ]          

Factor the first three terms in the parentheses  and simplify the last two terms

y =  (3/2) [ (x + 4)^2  + 4/3 ]       

Distribute the (3/2)  over the terms in the parentheses

y  = (3/2) (x + 4)^2  + 2    or  

(y - 2)  = (3/2) (x + 4)^2

Here is ONE standard form

3x^2+24x-2y+52    = 0       divide by 3

x^2 + 8x -2/3 y + 52/3         complet 'x' square

(x+4)^2    = 16    +2/3 y -52/3

    (x+4)^2 = 2/3 y -4/3

   (x+4)^2 =  2/3 (y - 2)

Let the radius  of the orange circle   = r

So......its area  =  pi * r^2

We can find the side length,s, of the  equilateral triangle thusly

tan (30°)  =  r / [(1/2)s]

1/ √3 = r / [ (1/2)s ]

(1/ √3) ( 1/2)s  = r

s =  (2√3)  r   = [ √12  ] r

And the area of the equilateral  triangle  is   (  √3/ 4 ) ([√12]r)^2  = 3√3 r^2

So....the blue area inside the equilateral triangle  = 

[area of equailateral triangle - area of small circle  ]  /  3 

[√3 r^2  - (1/3) pi r^2 ]  =   r^2 [ √3 - pi/3]      (1)

The radius  of the larger circle  can be found as 

√[ [√3r ]^2  + r^2 ]   =  √ [ 3r^2 + r^2 ]   =  2r

So....the area  of the  larger circle =  pi (2r)^2  =  4pi r^2

So   the area between the side of the equilateral triangle and the larger circle   is

[Area of larger circle  - area of equilateral triangle  ]  / 3  =

[ 4pi r^2  - 3√3r^2 ] / 3  =  r^2 [ (4/3)pi   - √3]    (2)

So the sum of  (1) and (2)   is  the sum of the blue areas

r^2  [ √3 - pi/3  + (4/3) pi - √3 ]   = pi r^2

So.....the orange and blue areas are equal   !!!!

Hi Daewei,

yes I fully understand this now...thank you for your input...

Hi Melody,

I always thought that after a log, everything following is part of that log function....I have learnt now that only the immediate value following the log, is applicable to the log function, and all other terms really stand "loose' from that function...Thank you very much for teaching me something again today....I really do appreciate...

It depends on the original format of the question. If it was originally written as 2*log(x+2) then no, you can't treat the two like a normal interger until you get rid of the logarithm. If, however, it was written 2*log(x)+2, then the two is not a part of the logarithm and can be treated as a normal interger because it is one. Once you determine the right format, it's merely a matter of stacking the logs and solving algebraically.

Hope this helped!

-Daewei

unfortunately I need to retract what I said about conversion to spherical coordinate since the resulting function \(g\) is a function of two variables and it makes no sense to say that it is continuous on \([0,2\pi]\); on \([0,2\pi]\times[0, 2\pi] \)maybe, but not on a single closed interval. Even then the function may not be continuous since in the denominator of \(g\) both \(sin(\theta)\)and \(sin(\phi)\)appear as factors and the function is not defined at \((0,0)\). I tried to extend Vipul Naik's two dimensional claims to three dimensions and it did not quite work the same way. I will keep working on this since I am definitely interested in it myself.

There is a 25% probabililty it will rain on the weekend.

The answer is 20.

The smallest positive a is 100.

The smallest positive a is 20.

The smallest positive a is 45.

You almost have the answer. The fraction in the denominator of the complex fraction is a ratio of two homogeneous functions; the top has degree two and the bottom has degree 6. So the fraction has degree -6+2 =-4, so its limit is undefined (not infinite) and that makes the limit of your original expression undefined.

For a more rigorous way of showing the limit is undefined convert the function\(F(x, y,z)=\frac{(|x|+|y|+|z|)^2}{x^2y^2z^2}\)into spherical coordinates and take the limit as r goes to zero. When converted you would have  \(F= \frac{1}{r^4}g(\theta,\phi)=r^{-4}g(\theta,\phi)\), where \(g\) is continuous on the closed interval \([0,2\pi]\) and therefore bounded from both above and below and is independent of r. This means that the limit is not defined (see Vipul Naik's monogram "LIMITS IN MULTIVARIABLE CALCULUS" that can be downloaded at

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=10&ved=2ahUKEwjj6Maxi-nlAhVRba0KHfRpAPkQFjAJegQIARAC&url=https%3A%2F%2Ffiles.vipulnaik.com%2Fmath-195%2Flimitsinmultivarcalc.pdf&usg=AOvVaw0cfu6Wc6RAMVgW7qT68-Mw

There is nothing to solve Tom because there is no equal sign.  

2^(2x+3) -2^(x+2) -3 * 4

\(2^{2x+3}-2^{x+2}-3*4\\ 2^{2x+3}-2^{x+2}-12\\ 2^{x+2+x+1}-2^{x+2}-12\\ 2^{x+2}*2^{x+1}-2^{x+2}-12\\ 2^{x+2}(2^{x+1}-1)-12\\ 4*2^x(2*2^x-1)-12\\ 8*2^{2x}-4*2^x-12\\ 8*4^x-4*2^x-12\\ 4(2*4^x-2^x-3)\)

I am just mucking around.

I really do not know what the question wants.

Well you could have a toffee apple, that way you sort of get the best of both worlds :)

LOL Who dreamed up that question!

Lets see I know how to use Google.

True, true! Still, I probably am a health nut. 

It may also be helpful to take a look at prime numbers for this problem as well.

First, do you know what palindromes are? Palindromes are numbers that are the exact same when read from right to left or left to right. For example, 1001 is a palindrome since it reads 1001 from right to left and it also reads 1001 from left to right. 

Another hint: Think about how many digits those numbers from 1000 - 100000 have. Could you possibly think of a way to use that?