All Answers 356338 Answers

you forgot the negative with the 12

Math help

My attempt:

\(\begin{array}{|rcll|} \hline 2A &=& \large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \Big(1-2\sin(\phi)\Big)^2\ d\phi \\ &=& \large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \left( 1-4\sin(\phi)+4\sin^2(\phi) \right) d\phi \\ &=& \large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \left( 1\right)\ d\phi -\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } 4\sin(\phi)\ d\phi +\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } 4\sin^2(\phi) d\phi \\ &=& \large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } d\phi -4\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \sin(\phi)\ d\phi +4\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \sin^2(\phi) d\phi \\ &=& \left[\phi \right]_{\frac{\pi}{6}}^{\frac{5\pi}{6}} -4\left[-\cos(\phi)\right]_{\frac{\pi}{6}}^{\frac{5\pi}{6}} +4\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \sin^2(\phi) d\phi \\ &=& \left[\phi \right]_{\frac{\pi}{6}}^{\frac{5\pi}{6}} +4\left[\cos(\phi)\right]_{\frac{\pi}{6}}^{\frac{5\pi}{6}} +4\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \sin^2(\phi) d\phi \\ &=& \dfrac{5\pi}{6}-\dfrac{\pi}{6} +4\left[ \underbrace{\cos\left(\frac{5\pi}{6}\right)-\cos\left(\frac{\pi}{6}\right) }_{=-\sqrt{3}} \right] +4\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \sin^2(\phi) d\phi \\ &=& \dfrac{2\pi}{3} -4\sqrt{3} +4\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \sin^2(\phi) d\phi \\ \hline \end{array} \\ \begin{array}{|rcll|} \hline \large{ \int \sin^2(\phi) } d\phi \\ &=& -\cos(\phi)\sin(\phi)+ \large{ \int } \cos^2(\phi)\ d\phi \\ &=& -\cos(\phi)\sin(\phi)+ \large{ \int } 1-\sin^2(\phi)\ d\phi \\ &=& -\cos(\phi)\sin(\phi)+ \large{ \int } 1\ d\phi-\large{ \int } \sin^2(\phi)\ d\phi \\ 2\large{ \int \sin^2(\phi) } d\phi &=& -\cos(\phi)\sin(\phi)+ \phi \\ \mathbf{\large{ \int \sin^2(\phi) } d\phi} & \mathbf{=}& \mathbf{\dfrac12\Big(\phi-\cos(\phi)\sin(\phi) \Big) } \\ \hline \end{array} \\ \begin{array}{|rcll|} \hline 2A &=& \dfrac{2\pi}{3} -4\sqrt{3} +4\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \sin^2(\phi) d\phi \\ && \quad \mathbf{\large{ \int \sin^2(\phi) } d\phi=\dfrac12\Big(\phi-\cos(\phi)\sin(\phi) \Big) } \\ &=& \dfrac{2\pi}{3} -4\sqrt{3} +2\Big[ \phi-\cos(\phi)\sin(\phi) \Big]_{\dfrac{\pi}{6}}^{\dfrac{5\pi}{6}} \\ &=& \dfrac{2\pi}{3} -4\sqrt{3} +2\Big[ \phi-\dfrac12\sin(2\phi) \Big]_{\dfrac{\pi}{6}}^{\dfrac{5\pi}{6}} \\ &=& \dfrac{2\pi}{3} -4\sqrt{3} +2\Big[ \dfrac{5\pi}{6}-\dfrac12\sin\left(\dfrac{10\pi}{6}\right)-\left(\dfrac{\pi}{6}-\dfrac12\sin\left(\dfrac{2\pi}{6}\right)\right) \Big] \\ &=& \dfrac{2\pi}{3} -4\sqrt{3} +2\Big[ \dfrac{2\pi}{3}-\dfrac12\sin\left(\dfrac{5\pi}{3}\right)+\dfrac12\sin\left(\dfrac{ \pi}{3}\right) \Big] \\ &=& \dfrac{2\pi}{3} -4\sqrt{3}+2\left( \dfrac{2\pi}{3} + \dfrac{\sqrt{3}}{2} \right) \\ &=& \dfrac{6\pi}{3} -3\sqrt{3} \\ &=& 2\pi -3\sqrt{3} \\ &=& 1.08703288447 \\ A &=& \dfrac{1.08703288447 }{2} \\ \mathbf{A} & \mathbf{=} & \mathbf{0.54351644224} \\ \hline \end{array}\)

simplify

\(\mathbf{\large{\sqrt[3]{250x^6}}}\)

\(\begin{array}{|rcll|} \hline && \mathbf{\large{\sqrt[3]{250x^6}}} \\ &=& \sqrt[3]{250}\sqrt[3]{x^6} \\ &=& \sqrt[3]{250}\cdot x^{\frac{6}{3}} \\ &=& \sqrt[3]{250}\cdot x^{2} \\ &=& \sqrt[3]{2\cdot125}\cdot x^{2} \\ &=& \sqrt[3]{2}\sqrt[3]{125}\cdot x^{2} \\ &=& \sqrt[3]{2}\sqrt[3]{5^3}\cdot x^{2} \\ &=& \sqrt[3]{2}\cdot 5^{\frac{3}{3}} \cdot x^{2} \\ &=& \sqrt[3]{2}\cdot 5^{1} \cdot x^{2} \\ &\mathbf{=}& \mathbf{\sqrt[3]{2}\cdot 5 \cdot x^{2} } \\ \hline \end{array}\)

The area of the ellipse is the area of the original circle multiplied by the determinant of matrix A.

I'll leave you to do the number crunching.

SUVAT Equations:

Thank you, CPhill !

See here : https://www.desmos.com/calculator/1lybspzeax

x = -3

x = 1

No table is shown   !!!!

1. The blue graph is the red graph shifted up 3 units

So...its equation is  y = f(x) + 3

2. The blue graph is the red graph shifted to the left by 2 units

So....its equation is   y =  f (x + 2)

2. This one seems harder than it really is

This function is odd [ it is symmetric about the orgin ] which means that if   ( - a, - b)  is on the graph....then so is (a, b)

So  u (-2.72)  leads to some - b

And u (2.72) leads to some b

So adding these together, we get  -b + b = 0

The same logic will apply to u(-0.81)  and u(0.81)

Adding their outputs will = 0

So

u(-2.72)+u(-0.81)+u(0.81)+u(2.72)  =    0

1.   h(7)  means that we go to 7 on  x and look at the y value at that point...this is 5

So   h(7) = 5

So...now we are trying to find h(h(7)) = h(5)

So now...we do the same thing.....go to  5 on x  and find the y value at that point = this is -1

So   h(h(7)) = h(5)  =   -1       !!!

1. The  "f( x - 3)"  shifts the the graph of f(x) to the right by 3 units

So...the vertex of f(x - 3)  is  ( 4, - 2)

2.  If the graph has an axis of symmetry of x = 5    and  (8, - 7) is on the graph.we added 3 to 5 to get (8,-7)...then....we must subtract 3 from 5 to get :

(5 - 3, -7) = (2, -7) is also on the graph

Note that the y coordinate is unaffected  !!!

Here you go :  https://www.desmos.com/calculator/4udsmi0nde

The solutions are shown

No prob !!!

I'm assuming that we have

y = √[x^2 + 8x ]       square both sides

y^2 =  x^2 + 8x        complete the square on x

y^2 =     x^2 + 8x + 16  - 16   factor

y^2 = ( x + 4)^2  - 16      add 16 to both sides

y^2 + 16 =  (x + 4)^2      take the positive root

√[ y^2 + 16 ] = x + 4       subtract 4 from both sides

√[ y^2 + 16] - 4 = x       swap x and y

√ [ x^2 + 16 ] - 4 = y   = the inverse    [ where x is also > 0 ]

thank you so much 

The two roots are  (-8,0)  and (7,0)

So....we have this form

y = a ( x + 8) ( x - 7)

And since (2, 100) is on the graph, we can find "a" thusly :

100 = a(2 + 8) ( 2 - 7)

100 = a (10)(-5)

100 =   -50a

100/-50 = a = - 2

So...the parabola is

y = -2 [ (x + 8) (x - 7) ]

y = -2 [ x^2 + x - 56 ]

y = -2x^2 - 2x + 112

Question 2

The domain  is  [-5, inf )

However.....from a positive square root we can only get 0  or a positive 

So.....the range is  [0, inf)

here's a graph to show this : https://www.desmos.com/calculator/quxajtdn9b