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sin v = 0,5

Decide the value of: sin 2v

\(\begin{array}{rcl|rcl|rcl} \sin(2v) &=& 2\cdot \sin(v)\cdot \cos(v) & \cos(v) &=& \sqrt{1-\sin^2(v)} & \sin(v) = \frac12 \\ & & & &=& \sqrt{1-\left(\frac12 \right)^2} \\ & & & &=& \sqrt{1- \frac14} \\ & & & &=& \sqrt{\frac34} \\ & & & &=& \frac{\sqrt{3}}{2} \\ &=& 2\cdot \frac12\cdot \frac{\sqrt{3}}{2} & &=& \frac{\sqrt{3}}{2} \\ &=& \frac{\sqrt{3}}{2} \\ \end{array} \)

\(\sin(2v) = \frac{\sqrt{3}}{2}\)

I don't understand what you are trying to show people Asad ..... :/

"How many joules of PE does a 12 kg rock have after falling from a 76 m cliff and reaches 7 m/s?"

E = PE + KE    E is total energy, PE is potential energy, KE is kinetic energy.

Top of cliff:  E = 12*9.8*76 + 0 joules

E stays constant so when velocity is 7 m/s

12*9.8*76 = PE + (1/2)*12*7^2

PE = 12*9.8*76 - (1/2)*12*7^2 or PE ≈ 9314 joules

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The angle is 120° to end up in quadrant II.  i.e. tan(120°) = -sqrt(3)

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Find b: 5a/c - y - d/b =z

\(\color{red}\frac{5a}{c}-y-\frac{d}{b}=z\\-\frac{d}{b}=z+y-\frac{5a}{c}\\-\frac{d}{b}=\frac{cz+cy-5a}{c}\)

\(-\frac{b}{d}=\frac{c}{cz+cy-5a}\\\color {blue}b=\frac{cd}{5a-cz-cy}\)

Find m: 4n/ab - c/m=x

\(\color{red}\frac{4n}{ab}-\frac{c}{m}=x\\-\frac{c}{m}=x-\frac{4n}{ab}\\ -\frac{c}{m}=\frac{abx-4n}{ab}\\ -\frac{m}{c}=\frac{ab}{abx-4n}\)

\(m=\frac{abc}{4n-abx}\)

  !

Suppose q is an angle in the standard position whose terminal side is in Quadrant II and tan Ø= - square root of 3

\(arctan( -\sqrt{3})=-60°\)

\(The \ free \ limb \ of \ the \ angle -60 ° \ forms \ the \ angle \ {\color{red}150 °} \ in \ the \ third \ quadrant.\)

.                                                                                           incorrect

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Thanks Alan! 

\(The \ free \ limb \ of \ the \ angle -60 ° \ forms \ {\color{blue}the \ angle \ 180° -60° = \ 120 °} \ in \ the \ third \ quadrant.\)

An engineer estimates the angle of elevation to the top of the building to be 50°.
After moving 1.5 meters further away, the angle of elevation was 40°.
How high is the top of the building?

\(\begin{array}{|lrcll|} \hline (1) & \tan(50^{\circ}) &=& \frac{h}{x} \\ & x &=& \frac{h}{\tan(50^{\circ})} \\\\ (2) & \tan(40^{\circ}) &=& \frac{h}{1.5+x} \\ & \tan(40^{\circ}) &=& \frac{h}{1.5+\frac{h}{\tan(50^{\circ})}} \\ & \tan(40^{\circ})\cdot \left( 1.5+\frac{h}{\tan(50^{\circ})}\right) &=& h \\ & 1.5\cdot \tan(40^{\circ}) + h\cdot \frac{ \tan(40^{\circ}) } { \tan(50^{\circ}) } &=& h \\ & h-h\cdot \frac{ \tan(40^{\circ}) } { \tan(50^{\circ}) } &=& 1.5\cdot \tan(40^{\circ}) \\ & h\cdot \left(1- \frac{ \tan(40^{\circ} ) } { \tan(50^{\circ}) } \right) &=& 1.5\cdot \tan(40^{\circ}) \\ & h\cdot \left( \frac{ \tan(50^{\circ})-\tan(40^{\circ} ) } { \tan(50^{\circ}) } \right) &=& 1.5\cdot \tan(40^{\circ}) \\ & h &=& 1.5\cdot \left( \frac{ \tan(50^{\circ})\cdot\tan(40^{\circ}) }{ \tan(50^{\circ})-\tan(40^{\circ} ) } \right) \\ & h &=& 1.5\cdot \left( \frac{ 1.19175359259\cdot 0.83909963118}{ 1.19175359259-0.83909963118 ) } \right) \\ & h &=& 1.5\cdot \left( \frac{ 1 }{ 0.35265396142 } \right) \\ & h &=& 1.5\cdot 2.83564090981 \\ & h &=& 4.25346136471\ m \\ \hline \end{array}\)

The top of the building is 4.25 m high.

Simplify the expression

1. Reorder the terms

\({26^3}p\times3!\rightarrow{26^3}\times3!\times p\)

2. Evaluate the power

\({26^3}\times3!\times p\rightarrow17576\times6p\)

3. Calc the product

\(17576\times6p\rightarrow105456p\)

Answer

\(105456p\)

See? You call everyone "blarney"!

All im asking is for you to stop mocking those people even if they are idiots. Please.

Btw i already wrote a response, it disappeared somehow. I think i remember questions that i cant find and people with much less points than they had a day ago.

Why would I want to do that?

If you have specific, curious questions, I may answer or not. This is finals’ week; I don’t have much time for blarney, but I will make time to roast a tasty vole, or at least devour it raw. 

As in the outside is all letters like this?

X + Y = Number 

2X + Y = Number

Around 11 to 1

 the answer is -4

-8 - -4 = -4

then the miuns and the negative changes to a postive 

-8 + 4 = -4

Nope. He have cancer probably.

for me about 9:30

volume of a sphere = (4π/3)(radius³)

4.5π = (4π/3)(r³)

Divide both sides by (4π/3) .

4.5π / (4π/3) = r³

4.5π * 3/(4π) = r³

27/8 = r³

Take the cube root of both sides.

\(\sqrt[3]{\frac{27}{8}}=r \\~\\ \frac{\sqrt[3]{27}}{\sqrt[3]{8}}=r \\~\\ \frac{3}{2}=r\)

r = 3/2 centimeters

\(\frac{3x}{8z}\,*\,\frac{12}{6y}=\frac{3x\,*\,12}{8z\,*\,6y}=\frac{36x}{48yz}=\frac{3x}{4yz}\)     :)

There are 3 * 60 = 180 minutes in 3 hours

If Billy keeps on smokin 5 cigarettes a minute,

after 180 minutes he will have smoked 180 * 5 = 900 cigarettes

And since he had 200 cigarettes to start with,

he will have 200 - 900 = -700 cigarettes after 3 hours 

pretty late

A story of two enemies that changed their lives forever.

what is 8x to the negative 3 power

\((8x)^{-3}=\frac{1}{(8x)^3}=\frac{1}{512x}\)

  !

how do i do this

y=1/4x+1 2y=2x-4

\(y=\frac{1}{4}x+1\\2y=2x-4\)

\(2\cdot( \frac{1}{4}x+1)=2x-4\\\frac{1}{2}x+2=2x-4\\6=\frac{3}{2}x\)       [\( y=\frac{1}{4}x+1\) 

\(x=4\)

\(y=1+1\)

\(y=2\)

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Thanks Alan!   I correct that.