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Thanks bree, you too

Here's (2)

Yeah ikr it was hard

Idk that was hard

What happened?

Here's (1)

I am good!

Because (-1)^10 = (-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1) = (1)(1)(1)(1)(1)(1) = 1, and 1 is positive.

Linda School held a walk-a-thon. the team started walking at 9:20 in the morning. They walked for 10 hours and 50 minutes. What time was it when the teams finished walking?

9:20am + 10:50 hours = 8:10pm, or in the evening. Another way of doing it is:

9 + 10 =19 - 12=7pm

20m + 50=70m - 60m=1 hour + 10 minutes

7 + 1 + 10minutes=8:10 in the evening.

4^2=4*4=16.

4cos^2θ = 6 - 9sin θ

4(1 -sin^2 θ) = 6 - 9sinθ

4 - 4sin^2θ  = 6 - 9sin θ     re-arrange as

4sin^θ - 9sinθ + 2  = 0   factor as

(4sinθ - 1) (sin θ  - 2)  = 0

Set each factor to 0.........the second one will have no solution since  sinθ  = 2  is impossible

4sinθ  = 1

sinθ  = 1/4    

And this happens at about 14.48° + n*360°   and at about 165.52° + n360°.....where n is an integer

Possible derivation:
d/dx((4 x y)/(x^2+y^2))
Factor out constants:
  =  4 y (d/dx(x/(x^2+y^2)))
Use the quotient rule, d/dx(u/v) = (v ( du)/( dx)-u ( dv)/( dx))/v^2, where u = x and v = x^2+y^2:
  =  ((x^2+y^2) (d/dx(x))-x (d/dx(x^2+y^2)))/(x^2+y^2)^2 4 y
The derivative of x is 1:
  =  (4 y (-(x (d/dx(x^2+y^2)))+1 (x^2+y^2)))/(x^2+y^2)^2
Simplify the expression:
  =  (4 y (x^2+y^2-x (d/dx(x^2+y^2))))/(x^2+y^2)^2
Differentiate the sum term by term:
  =  (4 y (x^2+y^2-d/dx(x^2)+d/dx(y^2) x))/(x^2+y^2)^2
Use the power rule, d/dx(x^n) = n x^(n-1), where n = 2: d/dx(x^2) = 2 x:
  =  (4 y (x^2+y^2-x (d/dx(y^2)+2 x)))/(x^2+y^2)^2
The derivative of y^2 is zero:
  =  (4 y (x^2+y^2-x (2 x+0)))/(x^2+y^2)^2
Simplify the expression:
Answer: |  =  (4 y (-x^2+y^2))/(x^2+y^2)^2

Possible derivation:
d/dy((4 x y)/(x^2+y^2))
Factor out constants:
  =  4 x (d/dy(y/(x^2+y^2)))
Use the quotient rule, d/dy(u/v) = (v ( du)/( dy)-u ( dv)/( dy))/v^2, where u = y and v = x^2+y^2:
  =  ((x^2+y^2) d/dy(y)-y d/dy(x^2+y^2))/(x^2+y^2)^2 4 x
The derivative of y is 1:
  =  (4 x (-(y (d/dy(x^2+y^2)))+1 (x^2+y^2)))/(x^2+y^2)^2
Simplify the expression:
  =  (4 x (x^2+y^2-y (d/dy(x^2+y^2))))/(x^2+y^2)^2
Differentiate the sum term by term:
  =  (4 x (x^2+y^2-d/dy(x^2)+d/dy(y^2) y))/(x^2+y^2)^2
The derivative of x^2 is zero:
  =  (4 x (x^2+y^2-y (d/dy(y^2)+0)))/(x^2+y^2)^2
Simplify the expression:
  =  (4 x (x^2+y^2-y (d/dy(y^2))))/(x^2+y^2)^2
Use the power rule, d/dy(y^n) = n y^(n-1), where n = 2: d/dy(y^2) = 2 y:
  =  (4 x (x^2+y^2-2 y y))/(x^2+y^2)^2
Simplify the expression:
Answer: |  =  (4 x (x^2-y^2))/(x^2+y^2)^2

Hi ! :)

I think the Headmistress and the Queen have now caught up on their royal corresponance.    Phew.....

Thank you all for being so patient :)

Hi Solveit :)

You might find this video interesting :)

https://www.khanacademy.org/math/trigonometry/trigonometry-right-triangles/intro-to-the-trig-ratios/v/basic-trigonometry

4 - - 2 = 4 + 2 = 6

Thanks 

\(\tan A = \frac {\textrm{opposite}} {\textrm{adjacent}} = \frac {a} {b}.\\ \text{tangent=tan.}\)

Hey Bree,

you our too have an amazing day! :)

ahem, in my world, those things are incorrect

Pain Makes Me Braver

Fear Makes Me Wiser

Heartbreaker Makes Me Stronger

why doesnt square root of 2 divided by 2 equal 0,5? If you take both of them squared it makes 2/4...

[Sqrt(2) / 2] IS QUITE DIFFERENT from [sqrt(2) / 2]^2

=0.707106................................................    = [.707106...]^2 =1/2

partially differentiate \(F = \frac{4xy}{x^2+y^2}\)

\(F_x = F \cdot ( \frac{1}{x} - \frac{2x}{x^2+y^2} ) \\ F_y = F \cdot ( \frac{1}{y} - \frac{2y}{x^2+y^2} )\)

circle inversion (unit circle (R=1) )  and axis-rotation 90 degrees.

circle inversion: \(x' = \frac{R^2\cdot x}{x^2+y^2}\\ y' = \frac{R^2\cdot y}{x^2+y^2}\\ r'\cdot r = R^2\)