All Answers 356066 Answers

If the standard form is a + bi then the trigonometric form is r*(cos theta + i*sin theta)  where

r = sqrt(a^2 + b^2)  and  theta = arctan(b/a)

So, for 4 (squareroot) 3 - 4i  we have r = sqrt(16*3 + 16)  or r = 8

                                                    theta  = atan(-4/4*sqrt(3)) = -30°

 4 (squareroot) 3 - 4i → 8(cos(30°) - i*sin(30°))  

3+2 (5) - |-7|   =

3 + 10 - 7  = 

13 - 7  =

6

Here's another way to consider this :

l a l   =  √(a^2)       

If we let a = -8, we have

l a l   = l -8 l  =   √ [ (-8)^2 ]   = √ [64]   =   8  

If P=210e^.0069*20

P=210.e^.0069*20

P=210.(1.006924*20)

P=210. 20.138478

P=402.76956. This is the correct answer the way you have written down. If it were: 210e^(.0069*20), then the right answer would be:241.0748655......etc.

Don't understand your last question! How do you derive e?, you mean?

"e"   can be generated by this sum

∞

∑   (1 / n! )  ≈  e

n = 0 

For example.......the first ten  terms of this sum  is given by :

1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! + 1/7! + 1/8! +1/9!  ≈ 2.718281525573192239858906

Of course........the more terms, the better the approximation...........

The logic and evaluation is sound!  

If you think it is wrong you need to explain why.

@CPhill your answer is WRONG

[4/(x+3)] + [(x+5)/(3-x)]       get a common denominator of ( x +3) (3 -x)   and we have

[ 4 ( 3-x) ] +  [(x + 5)(x + 3)] / [ (x + 3) (3 -x) ]      simplify

[ 12 - 4x  + x^2 + 8x + 15) ] / [ 3x + 9 - x^2 -  3x ]   =

[ x^2 + 4x + 27] / [ -x^2 + 9]  =

[x^2 + 4x + 27] / [ 9 - x^2]

a^2 = 18^2 + 13^2 - 2(18)(13)cos(64)

So.....taking the positive square root of both sides

a = sqrt [ 18^2 + 13^2 - 2(18)(13)cos(64) ]    = about  16.966

Ah!. But INFINITY is NOT a number, but a concept!!!.

\(\infty\)

7/14 = 1/2

According to special angles table, cos \(60^{\circ}\) = \(\frac{1}{2}\)

\(\therefore cos^{-1}(\frac{7}{14})=60^{\circ}\)

5 mod 3 = (the remainder of \(5\div 3\))= 2

\(\quad \frac{75\times 74\times73\times72\times71}{5\times4\times3\times2\times1}\\ = \frac{75}{5\times3}\times\frac{72}{4\times2}\times73\times71\times74\\ = 5\times9\times73\times71\times74\\=17259390\)

THERE IS NO BIGGEST NUMBER POSSIBLE! WHY? BECAUSE YOU CAN ALWAYS ADD 1 TO IT!.

C

Well they look like tangents but if that is not known for sure then I guess the answer is C

Another possibility is 11 crunchy tacos and 5 soft tacos.

11*2.5 + 5*3 = 42.50

e^(sqrt(-1)pi)+1

Here is the maths for it :)

\(e^{\sqrt{-1} \pi }+1\\ =e^{i \pi }+1\\ =Cos\pi+isin\pi+1\\ =-1+i*0+1\\ =-1+1\\ =0\)

Hey, guest, since you think "Maths is the Best" why don't you join up.  We want people like you to be a recognised part of our community :)

By "13 thirds", do yo mean:13 x 1/3 =13/3? If so, then: 13/3 =4+1/3 =4 1/3

1.261E-5?

1.261E-5 = 1.26 * 10^-5 = \(\frac{1.26}{10^{5} } \) = 0.0000126

       1.261E-5= 0.00001261=1.261 x 10^-5

Exactly

IS IT 9.05 (I rounded it to 2 d.p)

20x5/30+10x4/4x20/35

\(\frac{20\times5}{30}+\frac{\frac{10\times4}{4}\times20}{35}\\ =\frac{20\times5}{30}+\frac{\frac{10\times\not{4}}{\not{4}}\times20}{35}\\ =\frac{20\times5}{30}+\frac{10\times20}{35}\\\)

You should be able to take it from there. :)