All Answers 358578 Answers

Wow! I never thought it would be that easy to solve it!

-15 * pi/180= -pi/12 = -.2618 R

This is my understanding of this:

Since these are co-functions, the angles add up to 90 degrees:

x - 30 + 2x = 90

3x =90 + 30

3x = 120 

x = 120 / 3 

x = 40 degrees.

Hastag math,thank you for the answer,but it is wrong,and you answered it on the link you showed.

5. This answer is wrong

Thank you Rom,

I do appreciate your time...

                      2x^2-2x

x+3  |  2x^3+4x^2-6x+1

           2x^3+6x^2

                   -2x^2-6x+1

                   -2x^2-6x

                                    1        Remainder is  ' 1'

The answer given is wrong.

In order to correctly find the solution the trig functions must all be expanded out and combined.

This results in a 4th degree polynomial in either sin(x) or cos(x) depending on how you expand things.

I was working on writing up the LaTex for the solution when the editor hiccupped on me and I lost it all.

See if you can reproduce it.  It's not difficult.  A couple of trig identities and some algebra.

Deleted

Yes'

so what is the answer sir?

I think we will all agree that the initial assumption of 1= 2  is untrue.

While it IS true that  1 * 0 = 2 *0    you would have to eliminate the ' 0 ' on both sides of the equation to prove 1=2....to do that would require that you DIVIDE BOTH SIDES OF THE EQUATION BY ZERO....and we all know that DIVISION BY ZERO IS NOT ALLOWED/DEFINED.

You already asked this question and you got an answer.

How did you got the idea to divide it by 100 .

Ok thanks Ginger, and thanks guest too :)

Hi Jonathana

Determine, without a calculator, which is greater 4^100 or 6^75

\(\begin{array}{|rcl|cl|} \hline && 4^{100} && 6^{75} \\ &=& 4^{4\cdot 25} &=& 6^{3\cdot 25} \\ &=& \left(4^4\right)^{25} &=& \left(6^3\right)^{25} \\ &=& 256^{25} &=& 216^{25} \\ \hline \end{array} \)

\(\text{So $4^{100}$ is greater than $6^{75}$ }\)

\(\text{to expand on this a tiny bit suppose we start with}\\ 0\cdot x = 0\cdot y\\ \text{being specific about the algebraic step you mention we multiply both sides }\\ 0^{-1} \cdot 0 \cdot x = 0^{-1}\cdot 0 \cdot y \Rightarrow 1\cdot x = 1\cdot y \Rightarrow x = y\\ \text{where }0^{-1} \text{ is the multiplicative inverse of 0}\\ \text{but as Guest pointed out }0^{-1} \text{ does not exist, and thus the fallacy}\)

x mod 100 is just the remainder of x divided by 100

\((99x + 100y + 101z) = \\ 100x - x+100y + 100z + z \\ 100(x+y+z) - x + z\\ \text{when we mod by 100 the first part goes away since it's a multiple of 100 which leaves}\\ 100(x+y+z) - x + z \pmod{100} = -x + z\)

\(\text{let }u = (a,c,e),~v = (b,d,f)\\ \text{we are given that }\\ u\cdot u + v\cdot v = 6\\ u\cdot v = 3\\ \text{and we are asked for max of }(u+v)\cdot (1,1,1)\)

\(\text{By Cauchy-Schwarz}\\ |(u+v)\cdot(1,1,1)|^2 \leq (u+v)\cdot (u+v) \times (1,1,1)\cdot (1,1,1) = \\ (u\cdot u + v\cdot v + 2u\cdot v)(3) = \\ (6+2(3))(3) = 36 \)

\(\text{so }(u+v)\cdot (1,1,1)\leq 6 \\ \text{with the maximum occurring at 6}\\ \text{thus the maximum of }a+b+c+d+e+f = 6\)

Can you please explain how did that modulus function worked .

PLEASE

\(99x+100y+101z = 2009 \\ \text{mod both sides by 100}\\ z-x = 9 \\ z=x+9\\ 99x + 100y + 101(x+9) = 2009 \\ 200x + 100y + 909 = 2009\\ 200x + 100y = 1100\\ 2x+y=11\\ y=11-2x\)

\(\text{Now }y>0 \text{ so }1\leq x \leq 5\\ \text{thus we have the following 5 triplets}\\ (x,y,z) = (1,9,10),~(2,7,11),~(3,5,12),~(4,3,13),~(5,1,14)\)

\(P[\text{Joe eats at least two different kinds of fruit}] = \\ 1-P[\text{Joe only eats 1 kind of fruit}] = \\ 1 - 3\cdot \left(\dfrac 1 3\right)^3 = \\ 1 - \dfrac 1 9 = \\ \dfrac 8 9\)

Wait.. if k=1, the discriminat is 100-4*1^2=96, which is not a perfect square... so the solutions wouldn't be rational...right?

Maybe because multiplying by 0 is not reversible, if you divide by 0 it is undefined.

You don't need to multiply all this out

Just look at the denominator.  It's not allowed to be zero.

\(d = (5k^2 - 25k) \cdot 7k^4 = \\ 35k^5(k-5)\\ \text{so }k\neq 0,~5\)

so (a) and (d)