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Hello what are you doing ?

5x (to the second power) + 3y (to the second power) - 20x + 24y = -53

I think!:

5x^2 +3y^2 - 20x + 24y +53 =0

Find the derivative of the following via implicit differentiation:

d/dx(f(x)) = d/dx(sin^2(2 x))

The derivative of f(x) is f'(x):

f'(x) = d/dx(sin^2(2 x))

Using the chain rule, d/dx(sin^2(2 x)) = ( du^2)/( du) ( du)/( dx), where u = sin(2 x) and ( d)/( du)(u^2) = 2 u:

f'(x) = 2 d/dx(sin(2 x)) sin(2 x)

Using the chain rule, d/dx(sin(2 x)) = ( dsin(u))/( du) ( du)/( dx), where u = 2 x and ( d)/( du)(sin(u)) = cos(u):

f'(x) = cos(2 x) d/dx(2 x) 2 sin(2 x)

Factor out constants:

f'(x) = 2 d/dx(x) 2 cos(2 x) sin(2 x)

Simplify the expression:

f'(x) = 4 cos(2 x) (d/dx(x)) sin(2 x)

The derivative of x is 1:

f'(x) = 1 4 cos(2 x) sin(2 x)

Simplify the expression:

f'(x) = 4 cos(2 x) sin(2 x)

Expand the left hand side:

Answer: |f'(x) = 4 cos(2x) sin(2x) =f'(x) = 2 sin(4x)

\((x-a)^2 + (y-b)^2 = r^2 \text{(equation of circle )}\)

where your:

a=-2      (x-a)^2=(x+2)^2

b=4        (y-b)^2=(y-4)^2

r=4         r^2=16

Here s your graph:

\(\pi=3.14 \\ A_{circle}= \pi r^2 \\A_{circle}=7^2 \times \pi \\ \frac{49 \pi}{6}=25.7 \text{ (aprox area of sector)}\)

The area of the WHOLE circle is  pi x r^2 

  pi(7)^2 =  49pi           ONE sector (out of 6)   is:        1/6 x 49pi    =  49/6  x pi    

Yes.  As long as you mean -(4!).

Factorials of negative numbers are not defined, so (-4)! isn't defined.

Ensure the calculator is in degree mode (its default) and enter M/60  where M is the number of minutes.

Here's a slightly different (but no quicker) way:

.

I think The woman got it as I have nvr heard of a trapeze that was a grandfather *imagines shrivelled up old man swinging left to right* NOPE so there are no grandsons or grandaughters YET so the closes found would be alexis's niece.

If in an AP,ratio of 4th term and 9th term is 1:3 ,find the ratio of 12th and 5th term?

\(\frac{a+3d}{a+8d}=\frac{1}{3}\\ 3a+9d=a+8d\\ d=-2a\\~\\ \mbox{Now find }\\ \frac{T_{12}}{T_5}\\ =\frac{a+11d}{a+4d}\\ =\frac{a-22a}{a-8a}\\ =\frac{-21a}{-7a}\\ =3\)

So the ratio is 3:1

Ther is probably a quicker way to do this.  ://

Mornin' CP.....How was Mardi Gras?        Very foggy here this morning. Supposed to be 60 degrees today....snow will melt away.     ~jc

x+1/x=25   x^2+1/x^2=?

\(x+\frac{1}{x}=25\qquad \qquad x^2+\frac{1}{x^2}=?\\~\\ (x+\frac{1}{x})^2=25^2\\ x^2+\frac{1}{x^2}+2=625\\ x^2+\frac{1}{x^2}=623\\ \)

Alas, alas,

My posts are of such poor quality. Tis a good thing I have you, Camelot's own wise and learned Historian.

Thee must guide me to a more boutiful path  where the fruit be more abundant and its flavour be so delectable that if be not of this world!

I shall rely on you to lead the way :)

Headmistress Melody

 Camelot Academy

  The Realms Centre for Superior Education.

Hallo Gast!

x+1/x=25

x² + 1 = 25x

x² - 25x + 1 = 0

x1;2 = 12.5 +- √(12.5² - 1)

x1 = 24.96

x2 = 0.0401

x^2+1/x^2=?

x² + 1/x² = (x⁴ + 1) / x²

Gruß asinus :- )

!

Solve for x:

3 x+3/7 = 12

Put each term in 3 x+3/7 over the common denominator 7: 3 x+3/7 = (21 x)/7+3/7:

(21 x)/7+3/7 = 12

(21 x)/7+3/7 = (21 x+3)/7:

(21 x+3)/7 = 12

Multiply both sides of (21 x+3)/7 = 12 by 7:

(7 (21 x+3))/7 = 7×12

(7 (21 x+3))/7 = 7/7×(21 x+3) = 21 x+3:

21 x+3 = 7×12

7×12 = 84:

21 x+3 = 84

Subtract 3 from both sides:

21 x+(3-3) = 84-3

3-3 = 0:

21 x = 84-3

84-3 = 81:

21 x = 81

Divide both sides of 21 x = 81 by 21:

(21 x)/21 = 81/21

21/21 = 1:

x = 81/21

The gcd of 81 and 21 is 3, so 81/21 = (3×27)/(3×7) = 3/3×27/7 = 27/7:

Answer: |x = 27/7

Press "2nd" key first, then "asin" key followed by = key.

x+1/x=25

Solve for x:

x+1/x = 25

Bring x+1/x together using the common denominator x:

(x^2+1)/x = 25

Multiply both sides by x:

x^2+1 = 25 x

Subtract 25 x from both sides:

x^2-25 x+1 = 0

Subtract 1 from both sides:

x^2-25 x = -1

Add 625/4 to both sides:

x^2-25 x+625/4 = 621/4

Write the left hand side as a square:

(x-25/2)^2 = 621/4

Take the square root of both sides:

x-25/2 = (3 sqrt(69))/2 or x-25/2 = -(3 sqrt(69))/2

Add 25/2 to both sides:

x = 25/2+(3 sqrt(69))/2 or x-25/2 = -(3 sqrt(69))/2

Add 25/2 to both sides:

Answer: |x = 25/2+(3 sqrt(69))/2                  or              x = 25/2-(3 sqrt(69))/2

 x^2+1/x^2=?

Simplify the following:

x^2+1/x^2

Put each term in x^2+1/x^2 over the common denominator x^2: x^2+1/x^2 = x^4/x^2+1/x^2:

x^4/x^2+1/x^2

x^4/x^2+1/x^2 = (x^4+1)/x^2:

Answer: |(x^4+1) / x^2

I almost forgot one...

This is a response to your previous unsatisfactory posts:

1) (2.4)

2) (-2.0)

3) (4,0)

y = ax² + bx + c

1)  4 = 4a + 2b + c

2)  0 = 4a - 2b + c

3) 0 = 16a + 4b + c

1-2) 4 = 4b

        b = 1

2-3)  0 = -12a - 6b

        0 = -12a - 6

        a = - 1/2

2)     0 = -2 - 2 + c

        c =  4

y = - 0.5x² + x + c

: - )   !

Maybe I can go look for some better ones...:) 

No offense...!

Hi Coldplay,

I did notice any ...

Except the tine one I inlcuded.

I rather liked anons definition of devision.      LOL

You need to recommend posts if you think that they are really good.

Just please be discerning.    :D

Sun 14/2/16

1) History posts provided by Camelot's very onn Royal Historian.  I Thank yea Mistress Coldplay   

http://web2.0calc.com/questions/lantern-for-what-is-happening#r4

2) Trig ratios with 30 and 60 degrees.    Melody

http://web2.0calc.com/questions/sin-30_1

3) Easy but tricky algebra      Thanks guest, asinus and Melody :)

http://web2.0calc.com/questions/x-1-x-25-x-2-1-x-2

4) Area of triangels compared.  Unexp[ected outcome.  Thanks Chris.

https://web2.0calc.com/questions/triangle-area_1

5) Solveit does a good job here showing how to generate an equation for a circle :     Nominated by CPhill

[The Desmos graph was a nice touch, too   ]

http://web2.0calc.com/questions/help_27779

6) Ratio of terms in an AP      Thanks Melody and Alan,      Thanks for the nomination CPhill    

http://web2.0calc.com/questions/if-in-an-ap-ratio-of-4-t-h-term-and-9-t-h-term-is-1-3-find-the-ratio-of-12th-and-5th-term

7) I liked this "first principles"  problem that I did :    CPhill talking about himself  LOL     Nice work Chris and guest :)

[ This one required the correct identity to get started, along with a little Algebra "trick" in the middle   ......!!!  ]

http://web2.0calc.com/questions/first-principal

8) Using logs to find big numbers.  Thans Dragonslayer554 and Melody

http://web2.0calc.com/questions/math_51077

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