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Coc(2x) = 1        x = pi or 0    (and 2pi)

Thank you for checking

Yes....62.87 is correct

That makes sense, i see it now.Thank You

What about cos2x-1=0? I'm only getting pi as a solution. 

-sqrt2 +2     =    2 - sqrt2      Third choice

−2-√-2

2+√2

2−√2

−2+√2

these are the only options ,I cant find which one matches with your response

2 cos = 1

cos (x ) = 1/2   

ARCOS(X) = x = arcos(1/2) = 60 degrees     or   300 degrees      ( pi/3 radians   or   5pi/6 radians)

I meant to respond to your response

2 (-sqrt2/2)  - 4 (-1/2) = -sqrt 2 +2

yes....

Density is usually in units of   kg/m^3   or perhaps   mg/cm^3

Volume is given in m^3 or perhaps cm^3

Mass would be given in kg or mg

If you multiply volume by density you will get mass

  example     m^3  * kg/m^3  = kg   

Dango.....we need the table....do you have it? 

F= -kx    where k = spring constant  x = stretch distance

F=ma = 2 kg x 9.8 m/sec^2 = 19.6 N

19.6 = - k (.0288 m)       k = - 680 N/m

1.)

20% more than the oringinal cost is 20% over 100% which is 120%. 120% is 1.2 as a decimal.

12 * 1.2 = 14.40

Subtract the cost

$14.40 - $12 = $2.4

2.)

Two necklaces cost $15 so divide that in two to get $7.5 each.

Use the same process as number one to get 8.625 or $8.63 and the profit is 1.125 or $1.13.

Nevermind I got it. (0.2,-2.4)

You shouldn't have any trouble with the perimeter.  4 applications of the distance formula and sum.

For the area one way or the other you're going to need the height of the trapezoid.  That's not so easy.

As far as I can tell you're going to need to use the formula for the distance between two parallel lines.

$\text{the top of the trapezoid has slope}\\ m=\dfrac{5-7}{6-3}=-\dfrac 2 3\\ (y_t-7)=-\dfrac 2 3 (x-3)\\ y_t = -\dfrac 2 3 x+9$

$\text{the bottom of the trapezoid has the same slope}\\ (y_b-5)=-\dfrac 2 3 (x+1)\\ y_b=-\dfrac 2 3 x +\dfrac{13}{3}$

$\text{The distance between two lines is}\\ d= \dfrac{|b_2-b_1|}{\sqrt{m^2+1}}$

$b_1=9,~b_2=\dfrac{13}{3},~m = -\dfrac 2 3\\ d = \dfrac{|9-\frac{13}{3}|}{\sqrt{\dfrac 4 9 + 1}} = \dfrac{\frac{14}{3}}{\sqrt{\dfrac{13}{9}}}=\\ \dfrac{14}{\sqrt{13}} = \dfrac{14\sqrt{13}}{13}$

$\text{the length of the top segment is}\\ \ell_t = \sqrt{(6-3)^2+(5-7)^2} = \sqrt{13}\\ \text{the length of the bottom segment is }\\ \ell_b = \sqrt{(-1-5)^2 + (5-1)^2}=\sqrt{52} = 2\sqrt{13}$

$\text{Now we can apply the area of a trapezoid formula}\\ A= \dfrac{\ell_t + \ell_b}{2} h = \\ \dfrac{\sqrt{13}+2\sqrt{13}}{2}\times\dfrac{14\sqrt{13}}{13} = \\ \dfrac{3\sqrt{13}}{2}\dfrac{14\sqrt{13}}{13} = 21$

I kind of rushed through this and there may be an easier method.  I'm sure folks will chime in pointing out my errors.

Its actually

10a-7a=56-27+10

3a=39

a=13

I only got 2/6 i need proof that links to circle theorums

$f(g(x) = g(f(x))\\ f(g(x) = 3(ax+b)+2 = 3ax + (3b+2)\\ g(f(x)) = a(3x+2)+b = 3ax + (2a+b)\\ 3b+2 = 2a+b\\ 2b+2=2a\\ b=a-1$

$ab=20\\ a(a-1)=20\\ a^2-a-20=0\\ (a-5)(a+4)=0\\ a=5 \vee a=-4\\ 5+(-4)=1$