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sin ( arcsin (1/2)  + arrcos(2/3) )

arcsin (1/2)  = A

So

sin A  = 1/2      and cos A  =  sqrt ( 2^2 - 1^2) / 2 =  sqrt (3) / 2

arccos (2/3)  = B

So

cos B  = (2/3)    and sin B =   sqrt  ( 3^2 - 2^2) / 3  =  sqrt (5) /3

sin (A + B)   =   

sin Acos B  +  sin B cos A   =

(1/2)(2/3)  + (sqrt (5) / 3) (sqrt (3) / 2)   =

[ 2 +  sqrt (15)  ]  /  6

You can use this formula for multiplying:  r₁cis(A₁) · r₂cis(A₂)  =  (r₁·r₂)cis(A₁ + A₂).

In your problem, you are correct in multiplying the two constants, but you should have added the angles, not multiplied them.

x = 4 - t   ⇒   t  =  4 - x        (1)

y = (t - 1)^2       (2)

Put  (1)  into (2)

y  = ( 4 - x - 1)^2

y = ( 3 - x)^2

y = ( x - 3)^2

y = a ( x -3)^2  +  0

In the  form   y  = a(x - h)^2  + k       

a = 1     and the  vertex  is  (h,k)  = (3,0)

The initial point is  when t = -2

x  = (4 - -2)   = 6         y =  (-2 - 1)^2  = (-3)^2    = 9  ⇒   (6,9)

When t  = 2

x = (4 - 2) =  2          y  =  (2 - 1)^2   = 1^2   =  1    ⇒  (2, 1)

Since  (6.9)  is to  the right of (2,1)   the motion is  right to left

hmmm I think CPhill is correct...

If you plug in 2 to the second equation, you get: \(2^2+3(2)-38=4+6-38=\boxed{-28}\)

And \(-28 \neq 2\)

If you are dividing, use this formula:  r₁cis(A₁) / r₂cis(A₂)  =  (r₁/r₂)cis(A₁ - A₂)

\(\phantom{=\qquad}(\ 3\sqrt6\ \text{cis}(\frac{\pi}{8})\ )(\ 2\sqrt5\ \text{cis}(\frac{7\pi}{6})\ )\\~\\ {=\qquad}6\sqrt{30}(\ \text{cis}(\frac{\pi}{8})\ )(\ \text{cis}(\frac{7\pi}{6})\ )\\~\\ {=\qquad}6\sqrt{30}(\ \cos(\frac{\pi}{8})+i\sin(\frac{\pi}{8})\ )(\ \cos(\frac{7\pi}{6})+i\sin(\frac{7\pi}{6})\ )\\~\\ {=\qquad}6\sqrt{30}(\ \cos(\frac{\pi}{8})\cos(\frac{7\pi}{6})\ +\ i\cos(\frac{\pi}{8})\sin(\frac{7\pi}{6})\ +\ i\sin(\frac{\pi}{8})\cos(\frac{7\pi}{6})\ -\ \sin(\frac{\pi}{8})\sin(\frac{7\pi}{6}) \ )\\~\\ {=\qquad}6\sqrt{30}(\ \cos(\frac{\pi}{8})\cos(\frac{7\pi}{6})\ -\ \sin(\frac{\pi}{8})\sin(\frac{7\pi}{6}) \ +\ i\cos(\frac{\pi}{8})\sin(\frac{7\pi}{6})\ +\ i\sin(\frac{\pi}{8})\cos(\frac{7\pi}{6}) \ )\\~\\ {=\qquad}6\sqrt{30}(\ { \color{blue} \cos(\frac{\pi}{8})\cos(\frac{7\pi}{6})\ -\ \sin(\frac{\pi}{8})\sin(\frac{7\pi}{6}) } \ +\ i {\color{purple}(\ \cos(\frac{\pi}{8})\sin(\frac{7\pi}{6})\ +\ \sin(\frac{\pi}{8})\cos(\frac{7\pi}{6}) \ )} \ )\\~\\ {=\qquad}6\sqrt{30}(\ { \color{blue} \cos(\frac{\pi}{8}+\frac{7\pi}{6}) } \ +\ i\ {\color{purple} \sin(\frac{\pi}{8}+\frac{7\pi}{6}) } \ )\\~\\ {=\qquad}6\sqrt{30}(\ { \cos(\frac{31\pi}{24}) } \ +\ i\ { \sin(\frac{31\pi}{24}) } \ )\\~\\ {=\qquad}6\sqrt{30}\ \text{cis}(\ \frac{31\pi}{24}\ ) \)

It is vitally important to use a diagram, I would say. Luckily, I have taken the time to create one for you so that you can follow along. I have put all the information we know into the diagram below:

In order to find the perimeter, we need to express the length of each side of \(\triangle EFC\) in terms of x. \(\angle B\) is a right triangle, so by Pythagorean's theorem, \(AC=10\) and \(EC = 10-x\).

\(\triangle ACB \cong \triangle ECF \) by Angle-Angle Similarity Postulate, and we will use this similarity in order for us to express the remaining side lengths in terms of x like so:

\(\frac{10-x}{10}=\frac{EF}{8}\quad\quad\quad\frac{10-x}{x}=\frac{CF}{6}\\ EF = 8-\frac{4}{5}x\quad\quad CF=6-\frac{3}{5}x\) 

Now, find the perimeter by taking the sum of all the sidelngths of the desired triangle.

\(P=EC+EF+CF\\ P=(10-x)+(8-\frac{4}{5}x)+(6-\frac{3}{5}x)\\ P=24-\frac{12}{5}x\)

[x + y]  / 2   = √2 * √(xy)        square both sides

[ x^2 + 2xy + y^2 ] / 4  = 2 (xy)

x^2  + 2xy  + y^2  =  8 (xy)

x^2  + y^2  =  6(xy)      divide through by   x^2

1  + (y^2) / (x^2)  =  6(y/x)       rearrange  as

(y^2/x^2)  - 6(y/x)  +  1  =   0

(y/x)^2  - 6(y/x) + 1  =  0

Let   (y/x)   =  a

a^2   - 6a  + 1    =  0

a^2  - 6a    =   -1       complete the square on a

a^2 -6a + 9  = -1 + 9

(a - 3)^2   = 8       take both roots

a  -3   =  ±√ 8

a - 3  =  ±2√2

a =  3 ± 2√2

(y/x)  =    3 ± 2√2

Because the ratio of the width and the length is 4:3. The length is 4/3 times the width. If you calculate everything in terms of the width you get (4/3)w*1w = (4/3)w^2 = 1323 multiply both sides by 3/4 so it becomes, w^2 = 992.25. The square root of 992.25 is 31.5 so now we know the width of the rectangle. The length is 4/3 times as much as the with so 4/3*31.5 = 42 is the length. (31.5 + 42)*2 = 147

umm I think you need to say how many girls and boys there are... :)

x = y²/1-y 

The base and top of the prism are hexagons so the answer is B, a hexagon.

the answer is 126 because in the end you are going to have to move 9 steps in all. So, 9 choose 5 equals 126.

you calculate it without replacement because you are drwing them at the same time not 1 after the other.

you forgot togive us the number of girls and boys you might have tried to directly copy and paste

x  =  t - 2

y  =  2t² + 4

Let's solve the first equation for  t  and substitute that in for  t  into the second equation.

x  =  t - 2     Add  2  to both sides of this equation.

x + 2  =  t

Since  t  =  x + 2  we can substitute   x + 2  in for  t  into the second equation.

y  =  2t² + 4

y  =  2(x + 2)² + 4

I'll let you expand the right side of that equation if necessary.

To find the smallest possible value of  x,  plug in the smallest possible value of  t  into the equation   x = t - 2

And to find the biggest possible value of  x, plug in the biggest possible value of  t  into the equation   x = t - 2

When   t  =  -3,     x   =   -3 - 2   =   -5

When   t  =   1,     x   =    1 - 2   =   -1

So  x  is in the interval  [-5, -1]

not 100% sure on this, but I belave that the answer is B

We  have     [ ( -3 - (-1)  , (-2 - 6)  ]  =   [ -2 , - 8 ]

The  length of this veector  = ll a ll    =  √ [ ( -2)^2  + (-8)^2  ]  = √ [ 4 + 64 ]  =  √68  =

√4 * √17  =

2√17

r = √  [ (-3)^2 + (3√3)^2 ]  =  √ [ 9 + 27]  =  √36  =   6

θ = arctan (3√ 3 / -3)  = arctan  (-√ 3)    =    -pi/ 3  +  pi  =   2pi/3

z  =  6 cis  (2pi/3)

Since the point  (3, 6)  is on the graph of  y = g(x)  then  g(3)  =  6

And...

h(3)  =  ( g(3) )²   =   (  6  )²   =   36

Since  h(3) = 36  then the point  (3, 36)  is on the graph of  y = h(x)

3 + 36  =  39

It's not about hearts. It's about your dedication, ability, and quality of answers. You could have 10 times more than melody or cphill but if they are incorrect you still won't be moderator.

you earn the position of mod, you don't waltz into here and ask if you can be one.

Thank you! These were very helpful.

area of triangle   =   \(\frac12\)  *  base  *  height

We know the base is  4 inches.  To find the height we can use the Pythagorean theorem.

4²  +  h²   =   √[ 97 ]²

16  +  h²   =   97

h²   =   81

h   =   9

The height is  9  inches.

base   =   4 inches   =   4/12  feet   =   1/3  feet

height   =   9 inches   =   9/12 feet   =   3/4 feet

area of triangle   =   \(\frac12\cdot\frac13\cdot\frac34\ \)   sq feet   =   \(\frac18\)  sq feet

(x -3)^2  +y^2  =  9

( rcos θ  - 3)^2  + (rsinθ)^2   = 9

r^2cos^2θ - 6rcosθ + 9  + r^2 sin^2θ  = 9      subtract 9 from both sides

r^2 ( sin^2 θ  + cos ^2 θ)  - 6r cosθ  =  0

r^2 (1)   =  6r cosθ      divide out r

r  = 6cosθ

〈 -7, -3 〉  +  〈 -2, 10 〉   =   〈 -7 + -2, -3 + 10 〉   =   〈 -9, 7 〉