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\(13^{-1}\equiv 29 \pmod{47}\\ (29)(13) \equiv 1 \pmod{47}\\ (29)(47-34)\equiv 1 \pmod{47}\\ (29)(47)+(-29)(34)\equiv 1 \pmod{47}\\ -(29)(34) \equiv 1 \pmod{47}\)

\(34^{-1} \equiv (-29) \pmod{47}\\ (-29)\equiv (47-29) \pmod{47}\\ (-29)\equiv 18 \pmod{47}\\ 34^{-1} = 18 \pmod{47}\)

\(\text{as a check}\\ 18\cdot 34 = 612 = 13\cdot 47 + 1\\ 18 \cdot 34 \equiv 1 \pmod{47}\)

Using the Pythagorean Theorem.....the height of 1/2 of the cylinder is

√[5^2 - 3^2] = √[16]  =  4

So....the height of the whole cylinder =  8 units

The space between the sphere and the cylinder is given by :

Volume of sphere - Volume of cylinder  =

(4/3)pi (radius of sphere)^3  -  pi * (radius of cylinder)^3 * cylinder height  =

pi  [     (4/3)* 5^3    -  (3)^2 * 8 ]   =

pi  [   500 / 3  - 72 ]  =

pi [ ( 500 - 216) / 3 ]  =

pi  [ 284 / 3 ]

So

W  =    284 / 3

We can find the radius of the cone thusly :

Perimeter of cone  =  Perimeter of sector shown

2 pi * ( radius of cone)   =  (300 / 360) 2 pi * 18

radius of cone  =  (5/6) * 18

radius of cone =   15

The height of the cone is   √ [ 18^2 - 15^2 ]  =  √99 =  3√11

So....the volume of the cone is

(1/3)  pi * (radius)^2 * height  =

(1/3) pi ( 15)^2 * 3√11     =    

pi * 225√11

Dividing this by pi gives

225√11

\(88\)

x^2 - 6x + y^2 + 4y = 12       complete the square on x, y

x^2 - 6x + 9    + y^2 + 4y + 4   = 12 + 9 + 4

(x - 3)^2  + (y + 2)^2  = 25

Since R lies on the positive y axis, it must have the coordinates  (0, a)  where a is positive

So......x = 0    and we have

(0 - 3)^2 + ( y + 2)^2  = 25

9   + ( y + 2)^2  =  25     rearrange as

(y + 2)^2 =  16      take the positive root

y + 2  =  4

y = 2

So....R = (0, 2)

Here's the graph, YEEEEEET   :  https://www.desmos.com/calculator/j8cvgj36zi

Just find all the 2 digit factors. Factors are umbers that when divide into 144 and have a whole number answer.

Then add them up.

You can do this yourself.

6)  This forms an arithemetic series as follows:
First term =3994, Number of term =[10,002 - 12 / 10 + 1] =1,000 / 4 =250, Common difference =-16
Sum = N/2[2F + D(N-1)
Sum =250/2[2*3,994 + (-16*(249)
Sum = 125 [7,988 - 3,984]
Sum =500,500
Sum of 1 to 1,000 =[1,000 x 1001] / 2 =500,500

Does anyone have any  input for this one?

I have spent a lot of time looking.

If anyone else wants to step in, even with just thoughts, that would be good :)

Also

Does anyone know how to graph a cirlce (B) centred on the circumference of another cirlce (A) and moving on that arc?

(Like a planet going around the sun)

x is 1

1/3 (x+5)=2

x+5=2x3

x=6-5

x=1

thank you sooo much

\(\text{I assume this is (parens are your friend)}\\ \dfrac{x-\sqrt{x+1}}{x+\sqrt{x+1}}=\dfrac{11}{5}\)

\(\text{Let's pull a bit of a trick here}\\ a=\sqrt{x+1}\\ \dfrac{x-a}{x+a}=\dfrac{11}{5}\\ 5x-5a=11x+11a\\ 6x+16a=0\\ 3x=-8a = -8\sqrt{x+1}\\\)

\(9x^2 = 64(x+1)\\ 9x^2 - 64x-64 = 0\\ \text{Using the quadratic formula, (which I leave to you)}\\ x = 8,~-\dfrac{9}{8}\\ \text{Since we squared the equation we must check these roots}\)

\(x=8 \Rightarrow a = 3\\ \dfrac{8-3}{8+3} = \dfrac{5}{11} \neq \dfrac{11}{5}\\ \text{8 is a extraneous solution}\\ x=-\dfrac{8}{9}\Rightarrow a = \dfrac{1}{3}\\ \dfrac{-\dfrac{8}{9}-\dfrac 1 3}{-\dfrac{8}{9}+\dfrac 1 3} = \dfrac{11}{5}\\ \text{so }x=-\dfrac{8}{9} \text{ is an actual root, and the only one}\)

\(-\dfrac{1}{-\dfrac 8 9} = \dfrac 9 8\)

Thanks, CoolStuff........

I second your sentiments......!!!!

N=60
1 | 2 | 3 | 4 | 5 | 6 | 10 | 12 | 15 | 20 | 30 | 60 (12 divisors)
60 x 30 x 20 x 15 x 12 x 10 x 6 x 5 x 4 x 3 x 2 x 1 =46,656,000,000
                                                                       60^6=46,656,000,000

We  will  need to use Calculus to solve this.....

Looking at the circular cross section of the cylinder, it can be expressed as  a circle with a radius of 2 centered at the origin......its equation is :

x^2 + y^2 =  4 

 ....so.... 

y =  √ [ 4 - x^2 ]

The area of the circle covered by the oil is given by

Area of circle - area not covered by the oil  =

                     √3

pi * (2)^2  -    ∫   √[ 4 - x^2]  -  1   dx        ≈ 10.1096   ft^2

                    -√3

So...the volume of the oil is     

Area covered by oil * cylinder height  = 10.196 * 15   =  151.644 ft^2

So....the height of the oil when the tank is upright can be found as

151.644 =   pi * (2)^2 * height         

151.644 =  4 pi * height        divide both sides by   4 pi

151.644 / [ 4 pi ]   = height ≈    12.1 ft

The old radius is  ..... (1/2)(15/16) = 15/32 inches

The old volume is  .....    pi  * ( 15/32)^2 * (1/16)  in ^3

And 4 times this volume is   pi * (15/32)^2 * (1/16) * 4   in^3  =  pi * (15/32)^2 *(1/4) in^3

The  new diameter is   1 + 1/8  in   =  9/8 inches.....so....the new radius is 1/2 of this = (9/8) (1/2) = 9/16 inches

So we have that

pi * (15/32)^2 * (1/4)  =  pi * (9/16)^2  * height            [ divide out pi ]

(15/32)^2 / 4   =  (9/16)^2 * height            [  divide both sides by (9/16)^2 ]

(15/32)^2 / [ 4 * (9/16)^2 ]  =  height   =    25 /144  in

Let the original radius = r    and the original height = h    

So....the original volume is pi * r^2 * h  =    1 *  pi * r^2 * h

So....decreasing the radius by 20% produces .80r

And  increasing the height by 25% produces 1.25h

So....the new volume is pi * (.8r)^2 * 1.25h  = (64/100) * (125/100) pi * r^2 *h =

(64/100) * (5/4)  pi * r^2 * h =  [16 * 5] / 100  * pi * r^2 * h  = (80/100) * pi * r^2 * h =    .8 pi * r^2 * h

"Cancelling" out the pi * r^2 * h...... the percent change in the volume is

 ( [ 1 - .8 ] / 1   x 100  ) %  =     ( .2  x  100) %   =     20%

*** deleted ***    misread Q

\(\text{Let's split the students up into three non-intersecting sets}\\ F=\{\text{students taking only french}\}\\ S=\{\text{students taking only spanish}\}\\ FS=\{\text{students taking both french and spanish}\}\\\)

\(|FS|=|F|+|S|-|F\cup S| = 18+21-25 = 14\\ |F|-18-14=4\\ |S|=21-14=7\)

\(\text{4 outcomes lead to Michael being able to write about both classes}\\ (F,S),~(FS,S),~(FS,F)~,(FS,FS)\\ \text{These outcomes are non-overlapping so the probability}\\ \text{of their union is the sum of their individual probabilities}\)

\(P[(F,S)] = \dfrac{\dbinom{4}{1}\dbinom{7}{1}}{\dbinom{25}{2}}=\dfrac{7}{75}\\ P[(FS,S)] = \dfrac{\dbinom{14}{1}\dbinom{7}{1}}{\dbinom{25}{2}}\dfrac{49}{150}\\ P[(FS,F)] = \dfrac{\dbinom{14}{1}\dbinom{4}{1}}{\dbinom{25}{2}}=\dfrac{14}{75}\\ P[(FS,FS)] = \dfrac{\dbinom{14}{2}}{\dbinom{25}{2}}=\dfrac{91}{300}\)

\(P[\text{M. can write about both schools}] = \dfrac{28+98+56+91}{300} = \dfrac{91}{100}\)

Bruh, seriously you're asking on this website for AOPS homework answers. Bruh

\(30=2^1 \cdot 3^1 \cdot 5^1\\ 30^4 = 2^4 \cdot 3^4 \cdot 5^4\\ \text{the number of divisors is equal to the product of 1 plus each exponent in the }\\ \text{prime factorization of a given number}\\ n = 5\cdot 5\cdot 5 = 125\)

Comet's second queston is actually this :

2)Solve for x over the interval of [0, 2pi]

-2 + tan^2x=0

Add 2 to both sides

tan^2x  =  2       taske both square roots

tan x = ± √2

We either have that    

tan x = √2       

Take the tan inverse to find the angle in  rads 

arcctan (√2)  ≈ .955 rads    this will also occur at   pi +  .955 rads = 4.096 rads

Also

tan x = -√2

arctan ( -√2)  ≈ -.955 rads     =   [2pi -  .955] rads  ≈ 5.328 rads

And once more at  [5.328 - pi] rads   ≈ 2.19 rads   

So....the four answers are ≈   [ .955 , 2.19, 4.096 and 5.328 ]   rads

2)Solve for x over the interval of [0, 2pi]

-2 tan^2x=0

Divide both sides by - 2

tan^2  x  = 0        take the square root and we get that

tan x  =  0

This happens at   x =   0 ,  pi  and 2pi    rads     on the requested interval

4cos(2 theta)+3=0     subtract   3 from both sides

4 cos (2 theta)  =  - 3     divide both sides by 4

cos (2 theta)  = -3 / 4

cos (theta)  =  -3/4    at  ≈ [ 2.42   ±  2pi ]  rads  and at ≈ [3.86 ± 2pi ]  rads

So

cos ( 2 theta) =   (  [ 2.42 ± 2pi ] / 2   )  rads   and   at   ( [ 3.86 ± 2pi ] / 2 rads )  =

[ 1.21 ± pi ] rads    and   [ 1.93 ± pi ]  rads

Aluminum molar weight  26.981 gm /mole

135 gm  / 26.981 gm/mole =5.00 moles

It takes TWO moles of aluminum per reaction   ( 2 Al  in the equation)

5 moles/ 2moles/rxn  x -1052 kj/rxn =  - 2631  kj