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Area of a square   = side^2   = [11 cm]^2   = 121 cm^2 

d. cos (2x) - cos x  = 0

2cos^2x - 1 - cos x   =0

2cos^2x  - cos x  - 1   =  0   factor

(2cos x + 1) (cos x - 1)  = 0      set each factor to  0  and solve for  x

2cos x  + 1  = 0

2cos x  = -1

cos x  = -1/2    and this happens at   2pi/3 + n*2pi    and  4pi/3 + n*2pi   where  n is an integer

And

cos x - 1   =  0

cos x   = 1    and this happens at   0 + n*2pi    where n is an integer

Well I am glad that you got it. 

Pity there was no 'thank you' thrown in there while you were at it though.

A 'thank you' is always appreciated.

c. 2sin^2 θ  + 3sin θ  = 2

2sin^2 θ + 3sinθ - 2  = 0   factor

(2sinθ - 1) (sin θ + 2)   =  0

Setting each factor to 0....solve for  θ

2sinθ - 1  =  0

2sinθ  = 1

sinθ  = 1/2     and this happens at    pi/6 + n*2pi   and   5pi/6 + n*2pi    where n is an integer

And

sin θ  + 2   =0

sinθ  = -2     [ this has no solution ]

Thanks very much guest, that is indeed the same question only not the desired answer.

There was another one more recent that that did not have a picture with it. It had a few different answers.

b. sinx  - cos^2 x - 1   = 0

sin x  -  [ 1 - sin^2x ]  - 1   = 0

sin^2x  + sinx  - 2  = 0     factor

(sin x - 1) ( sin x  + 2)  = 0 

Set each factor to 0 and solve for  x

sin x  - 1   = 0

sin x  = 1  and this happens at   pi/2 + n* 2pi   where n is an integer 

And

six + 2  =  0   has  no solution

a. sin^2x  = 2sin^2(x / 2)

Note  sin^2 (x / 2 )   =  1 - cos x

So we have

sin^2x  = 1 - cos x

1 - cos^2x  = 1 - cos x      subtract 1 from both sides

-cos^2x  = -cos x       multiply through  by  -1

cos^2x  = cosx

cos^2x - cos x  = 0

cosx ( cos x - 1)  = 0

Set each factor to 0  and we have that

cos x  =  0     and this happens at   pi/2  + n*pi    where n is an integer

and

cos x  -  1   = 0

cos x  = 1     and this happens at   0  + n* 2pi    where  n is an integer

f(x)  =x^3  - 4x^2  - 9x - 36   factors as   

x^2(x - 4) - 9(x - 4)   =

(x - 4) (x^2 - 9)  =

(x - 4) (x + 3) ( x - 3)

So only answers  2,3, and 4  are correct

f(x) = x^3 + 5x^2 +6x

A.....when x  = 2 we have  (2)^3 + 5(2)^2 + 6(2)   =  8 + 20 + 12  =  40...so....not true

B.....x^3 + 5x^2 + 6x  factors as  x  ( x + 3) ( x + 2)....so...  true

C...when x   - 2  we have  (-2)^3 + 5(-2)^2 + 6(-2)      =  -8  + 20 - 12   =  0.....so...true

D.....x - 2 is not a factor....so...f(x) divided by  x - 2  will not result in a remainder of  0 ...so....not true

Here are ALL the digits from 0 to 9 in 1 to 500:

(0=91, 1=200, 2=200, 3=200, 4=200, 5=101, 6=100, 7=100, 8=100, 9=100)

Oh nvrm I get it

Why do you multiply by 6? 

....further   y^13 =  y^3 y^3 y^3 y^3 y     so you can take all of those y^3 's out

   and   x^5 = x^3 x^2       so you can take the x^3 out

2 x y^4 \(\sqrt[3]{2x^2 y}\)

The exponent   3/7    means   raise everything to the  3 'power' and then take the '7th' root....

....so first choice is no good (cube root or 3rd root)

5^3 = 125  so the 3rd choice looks good

(x^3)^3 = x^9    so the third choice still looks good

(y^4)^3 = y^12      so the third choice it is!

eh i got nothing better to do

Sure mate. Why not?

because they are never right 

why are obtuse angles so depressed 

sure

I'm not really good at math lol

Ugh, I have to read from the table. Well, whatever, I'll try. For, City 2, we have \(7+17+20+19+7+2=72.\) And, the mean is just \(\frac{72}{6}=\boxed{12}.\)

one person: \(\frac{2}{8}\) of a pizza, or \(\frac{1}{4}\) of a pizza.

Since we have \(3\) pizzas, there will be \(\frac{3}{\frac{1}{4}}=3*4=\boxed{12}\) people.

Yes, it's in the form \(ax^2+bx+c\), and you plug the values into this formula: \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\). Neat that web2.0calc gives you the quadratic formula, when you press the latex button! Remember that, \(\pm\) is plus or minus.

You can break the original value into \(\sqrt[3]{16}\sqrt[3]{x^5y^{13}}.\) We know that \(\sqrt[3]{16}=2\sqrt[3]{2}\), because of the power,  \(2^4.\) Thus, the answer is \(\boxed{2\sqrt[3]{2}\sqrt[3]{x^5y^{13}}}\).

Maybe this is what Melody is referring to: https://web2.0calc.com/questions/a-square-defg-varies-inside-equilateral-triangle