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FV = PV x  e^RT

FV = 2,000 x e^(.05*10)

FV = 2,000 x e^0.5

FV = 2,000 x  1.648721271......

FV =$3,297.44

Not sure if this is meant to be  (1) y∝ √x or (2) y∝ 1/√x

If (1):   y = k*√x         7 = k√36      7 = 6k    k = 7/6      y = (7/6)*√4      y = (7/6)*2     y = 7/3

If (2):    y = k/√x         7 = k/√36     7 = k/6   k = 42      y = 42√4     y = 42*2     y = 84

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"A 2.4 kg block rests on a frictionless wedge that has an inclination of 36◦ and an acceleration to the left such that the block remains stationary relative to the wedge; i.e., the block does not slide up or down the wedge. The acceleration of gravity is 9.81 m/s 2 .

Find the magnitude of the acceleration of the block-wedge system. Answer in units of m/s 2 ."

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2sin(2x)/1-sin^2(x)=5 can convert to tan(x)=1,25

a) Show that the equation \(\tfrac{2\sin(2x)} { 1-\sin^2(x) } = 5\) can convert  to tan(x)=1,25

\(\begin{array}{|rcll|} \hline \dfrac{2\sin(2x)} { 1-\sin^2(x) } &=& 5 \quad & | \quad : 2 \\\\ \dfrac{\sin(2x)} { 1-\sin^2(x) } &=& \frac{5}{2} \\\\ \dfrac{\sin(2x)} { 1-\sin^2(x) } &=&2.5 \quad & | \quad \sin(2x) = 2\sin(x)\cos(x) \\\\ \dfrac{2\sin(x)\cos(x)} { 1-\sin^2(x) } &=& 2.5 \quad & | \quad 1-\sin^2(x) = \cos^2(x) \\\\ \dfrac{2\sin(x)\cos(x)} { \cos^2(x) } &=& 2.5 \quad & | \quad : \cos(x) \\\\ \dfrac{2\sin(x)} { \cos(x) } &=& 2.5 \quad & | \quad : 2 \\\\ \dfrac{\sin(x)} { \cos(x) } &=& 1.25 \quad & | \quad \dfrac{\sin(x)} { \cos(x) } = \tan(x) \\\\ \tan(x) &=& 1.25 \\ \hline \end{array}\)

b) Solv the ecvation tan(x)=1,25 fully

\(\begin{array}{|rcll|} \hline \tan(x) &=& 1.25 \quad & | \quad \arctan() \text{ both sides }\\ x &=& \arctan(1.25) + n\cdot \pi \quad & \quad n \in Z \\ x &=& 51.3401917459^{\circ} + n\cdot \pi \quad & \quad n \in Z \\ \hline \end{array}\)

*edit*

Sorry about this! I didn't pay good enough attention here...the question says to rotate about the square's vertical line of symmetry, which makes a cylinder with a radius of 7 cm. and a height of 14 cm. Thanks to Alan for bringing this to my attention!!

So... the volume  =  pi * 7² * 14  =  pi * 49 * 14  =  686pi  cubic cm

I'll leave the old answer below even though I think it is wrong.

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The rotation will form a cylinder with a radius of 14 cm and a height of 14 cm, like this....

volume of cylinder  =  (pi * radius²) * (height)

                               =  (pi * 14²) * 14

                               =  (pi * 196) * 14

                               =  2744pi

The volume is 2744pi cm³ .

(x - 1)(x + 2)  =  70

                                      Multiply out the left side of the equation.

x² + x - 2  =  70

                                      Subtract  70  from both sides of the equation.

x² + x - 72  =  0

Can you think of two numbers add to  1  and multiply to  -72?   →   -8  and  +9 .

So the left side can be factored like this...

(x - 8)(x + 9)  =  0           Now we set each factor equal to zero and solve for  x  .

x - 8  =  0      or      x + 9  =  0

     x  =  8      or            x  =  -9

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I don't know the best way to do the second one, but I'm sure someone else on here does.

Here's the third one:

9(x - 1)²  =  4(x - 2)²     Take the ± square root of both sides.

                                     (We only need to write the ± on one side though.)

\(\sqrt{9(x-1)^2}\,=\,\pm\sqrt{4(x-2)^2} \\~\\ \sqrt9\cdot\sqrt{(x-1)^2}\,=\,\pm\sqrt4\cdot\sqrt{(x-2)^2} \\~\\ 3(x-1)\,=\,\pm2(x-2) \)

\(3x-3\,=\,\pm(2x-4)\)       Split this into two seperate equations.

\(\begin{array} \ 3x-3=+(2x-4)\qquad\text{or}\qquad&3x-3=-(2x-4) \\~\\ 3x-3=2x-4&3x-3=-2x+4 \\~\\ x-3=-4&5x-3=4 \\~\\ x=-1&5x=7 \\~\\ &x=\frac75 \end{array}\)

On the calculator, you can make fractions by using the backslash to split between numerator and denominator

For example: 7/9 makes 

Note that     (a + b)(a - b)  =  a² - b²

So...           \(({\color{violet}\,19\sqrt{h}} + {\color{blue}4\sqrt{2h}})({\color{violet}\,19\sqrt{h}} - {\color{blue}4\sqrt{2h}})\,=\,({\color{violet}\,19\sqrt{h}})^2-( {\color{blue}4\sqrt{2h}})^2\)

\((\,19\sqrt{h})^2-( 4\sqrt{2h})^2 \\~\\ =\,19*19*\sqrt{h}*\sqrt{h}-4*4*\sqrt{2h}*\sqrt{2h} \\~\\ =\,361h-16*2h \\~\\ =\,361h-32h \\~\\ =\,329h\)

ques is

3x-0.5=5x-2

keeping algebric terms at one place..

2-0.5 = 5x-3x

1.5 = 2x

x = 1.5/2

x = 0.75

/Solved by onlinerechargetricks.com

Circle \;\;\Gamma\;\; \text{ is the incircle of }\triangle ABC \text{ and is also the circumcircle of }\triangle XYZ\text{ The point X is on }\overline{BC} \text{ point Y is on }\overline{AB}, \textand the point Z is on }\overline{AC}.\;\; If\;\; $\angle A=40^\circ, \;\;\angle B=60^\circ,\;\; and \;\;\angle C=80^\circ, \text{what is the measure of }\angle AYX\; ?

\(\text{Circle }\Gamma\;\; \text{ is the incircle of } \triangle ABC \text{ and is also the circumcircle of } \triangle XYZ\\ \text{ The point X is on }\overline{BC} \text{and the point Y is on }\overline{AB}, \\ \text{and the point Z is on }\overline{AC}.\\ If\;\; \angle A=40^\circ, \;\;\angle B=60^\circ,\;\; and \;\;\angle C=80^\circ, \text{what is the measure of }\angle AYX\; ? \)

   Let O be the centre of the circle.

  OX=OY                                 equal radio

  So

  OXY is an isosceles triangle

  120+2

  so

  < AYX =

I am sorry, this was a full answer but 3/4 of it has been deleted.

There is obviously a software problem.       

I shall report it as a problem :/

simplify | (19 sqrt(h) + 4 sqrt(2 h)) (19 sqrt(h) - 4 sqrt(2 h))
361h - 76sqrt(2)h + 76sqrt(2)h - 32h
361h - 32h 
329h

3x - 0.5 = 5x - 2   isolate x to the LHS

3x - 5x = -2 + 0.5

-2x = - 1.5             divide both sides by -2

x = -1.5 / -2

x =0.75

a*b=2a + 3b

a = -3 and b = 1

a = 0 and b = 0

a = 1 and b = -1

a = 2 and b = -4

a = 4 and b = 8

a = 5 and b = 5

a = 6 and b = 4

a = 9 and b = 3

You need to use brackets darkcalculus :)

23+9/7-3=8

should be

(23+9)/ (7-3)=8

Not if you do not ask a question :/

Rotate 90º Counter clockwise Around the Origin And then Reflect Over the X axis With (-8,0),(-11,-5),(-4,-10),(-6,-4)

5x6= 30 dude

OK, GR....!!!

THANK YOU CPhill SO MUCH

The area  is given by the side^2

So

200  = s^2        take the square root of both sides

√200  = s

√[ 100 * 2]   = s

10√2  = s  ≈  14.14 ft

Using the Law of Sines to find the angle opposite the side of 15, we have

arcsin [ 15*sin (108) / 28 ]  ≈  30.63°

So...the remaining agle must be   = 180 - 108 - 30.63   ≈  41.37°

So......applying the Law of  Sines once more, we have

  x  / sin (41.37)   = 28 / sin (108)      ......  multiply both sides by  sin (41.37)

x  =   28* sin( 41.37) / sin (108)  ≈ 19.46 

Not totally sure about this....but.....look at the following image : 

We can anchor Q and R  at two consecutive points on the circle.......  PQR  will have five possible values........

Let P  be positioned at P1 and PQR will can be an inscribed angle measuring (5/7) of 180°

Let P be positioned at P2 and PQR will be an  inscribed angle measuring (4/7)  of 180°

.......

Lastly.....let P  be positioned at P5 and PQR will  be an inscribed angle measuring (1/7) of 180°

Here's a suggestion if the fractions combined with decimals are messing with your mind.

\(\frac{3.2}{0.04}*\frac{100}{100}=\frac{320}{4}\)

Multiplying by 100/100 has no effect on the value of this fraction as it is equivalent of multiplying by 1. 

\(\frac{320}{4}=\frac{32*10}{4}=\frac{8*10}{1}=80\)

\(10(\frac{1}{6})+5(\frac{2}{6})+-6(\frac{3}{6}) \)

\(\frac{10}{6}+\frac{10}{6}-\frac{18}{6}\)

\(\frac{20}{6}-\frac{18}{6}\)

\(\frac{2}{6} = \frac{1}{3}\)