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Thank you Rosala.    

I wish you and everyone a fabulous New Year.  :)

If they are linear equations (no squared terms, no cubics, etc), there are seveal options. If you know how to use matrices, use them; otherwise, if you know how to use determinants, use them. If not, reduce four equations in four unknowns down to three equation in three unknowns.

Example:

3w - 4x + y - 2z = 7

w + 3x - 5y + 7z = -7

2w + 2x + 5y - 2z = 26

4w - 3z + 2y - z   = 16

Choose a variable to eliminate from all four equations. If you choose w:

Looking at equations 1 and 2:  

  3w - 4x + y - 2z  =  7                     --->    3w - 4x +   y -   2z  =  7 

w + 3x - 5y + 7z  =  -7     --->   x -3   --->   -3w - 9x + 15y - 21z  =  21

     Add down the columns:                               -13x + 16y - 23 z  =  28

Looking at equations 2 and 3:

w + 3x - 5y + 7z  =  -7     --->   x -2   --->   -2w - 6x +10y - 14z  =  14

2w + 2x + 5y - 2z = 26                        --->   2w + 2x + 5y - 2z   =  26

     Add down the columns:                                 -4x + 15y - 16z  =  40

Looking at equations 2 and 4:

w + 3x - 5y + 7z  =  -7     --->   x -4   --->   -4w - 12x + 20y - 28z  =  28

4w - 3z + 2y - z   = 16                       --->     4w - 3z +   2y -     z   = 16 

     Add down the columns:                                -15w + 22y - 29z  =  44

Putting these equations together:

      -13x + 6y - 23 z  =  28

      -4x + 15y - 16z  =  40

      -15w + 22y - 30z  =  44

Now you have three equation in three unknowns; can you solve it from here?

ask your mom

u'(t)+u(t)=k   ...this is a linear differential equation which is in the form  du/dt + p(t) u = g(t)......note that p(t)  = 1 

We need to multiply through by an integrating factor which is given by e^∫p(t)dt  = e^∫(1)dt = e^t

So we have

e^t u'(t) + e^t u(t)= ke^t

And because of our employment of the integrating factor, we can write the left side in the form of the product rule, thusly....

[ e^t u(t) ] '  = ke^t    ....integrate both sides

∫[ e^t u(t) ] ' dt  = ∫ ke^t dt

e^t u(t)  = ke^t + C    divide both sides by e^t

u(t) = k + Ce^-t

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Note that this solution "works" because

u'(t) = -Ce^-t   and u(t)  = k + Ce^-t

And their sum = k

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24 000 000 000 / 60 000 000

Cancelling multiples of ten in the numerator with corresponding multiples of ten in the denominator:

--->   24 000 / 60  =  2400 / 6  =  <your answer>

r  =  √[x² + y²]   --->   r  =  √[5² + 6²]   --->   √[61] 

θ  =  tan^-1(y/x)   --->   θ  =  tan^-1(6/5)   

(r, θ)  =  ( √(61), tan^-1(6/5) )

u'(t) + u(t) =  k

--->   u'(t)  =  k - u(t)

--->   ∫u'(t)dt  =  ∫[k - u(t)]dt

--->   u(t) + C₁  =  ∫kdt - ∫u(t)dt

--->   u(t) + C₁  =  kt + C₂ - ∫u(t)dt

--->   u(t)  =  kt - ∫u(t)dt + C

It is the side that is opposite the right angle in a right triangle.

Divide the mass of an average blue whale by the mass of an average giraffe.

Thanks and same to you

sorry -π/4 ?

but if it were -π/2 ? plz

Lol ty it was more of a joke from me, but ty anyway xD

cos(π/2) = 0   --->   cos²(π/2) = 0   --->   -3cos²(π/2) = 0

2x^2-4xy   ....taking out the greatest common factor, 2x, we have

2x (x - 2y)

And that's it......!!!!

5^25 > 25^5     well...let's see......we can write

5^25  ???   (5^2)^5 

5^25 ??? 5^10

And

5^25 > 5^10

So.....it's true...!!!

A formula for the volume of a cone is:  V  =  (1/3) · π · r² · h

Since you know the values for V and h, can you solve this formula for r?

If so, multiply that answer by 2 to get the diameter; if not, post back and we can help you.

Happy New Year DavidQD

Happy New year CPhill

and

Happy New Year NotTheBestMathMaster   :)

Happy New Year, DavidQD.....!!!

Now that I think about it . . . I’ve been gone less than a “blink”

Happy new year everyone!

HI Melody and CPhill !!!

It seems I’ve been gone for a cosmic fortnight -- not quite as long as a decade but longer than an imperial minim …

Just because my answer isn't as mind boggling doesn't mean I dont deserve a Thumbs Up

Everyone pays no attention to MY answer :P