All Answers 359202 Answers

Annie - 300
Bobby - 280
Ciro - 280
Denise - 280.1

Annie:   300

Bobby:  280

Ciro:     280

Denise: 280.1

Sum = 1140.1

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Nice answer Zac  

yay...thanks titanuim rome!

Melody, it is the "mg" vector that constitutes the hypotenuse of the right-angled triangle, not the  "mgsinα" line  (I didn't attempt to draw the two magnitudes to scale; I probably should have done!).

To convince yourself that the component along the slope is mgsinα think of the limit when α = 0.  The component along the "slope" must then be zero.

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Thanks Mathcad    

auctually, i give them like 5 points each time LOL

Why? Did CPhill put the mower over the Roman Zero?     

You did not answer my question for number 11 MG

The question was write this correctly with brackets and NO latex.

$$\frac{5x+2}{6y}\times9xy^2$$

Q7) DOH!

I thought it was times, otherwise i would have done what you done 

Q9) Thanks for the explanation :D , i thought the answer had be divided back .

Q11) Sorry about that, i forgot about it and how text can be easily confused wih something else.

$$\\5x+\frac{2}{6}*y*9xy^2=5x+\frac{2y}{6}*9xy^2\\\\$$

Well thejamesmachine, that is really cool,  just don't get too attached to them.  LOL

Thanks Alan but I am having problems with this :/

I can see that the two angles labelled theta in the first my pic are equal.

and

i can see using the second theta in my pic that

sin(theta)=weight down slope/mg      so

weight down slope = mg*sin(theta)

BUT

If I use the same angle which you have labeled alpha then

sin(alpha)=mg/weight down the slope

weight down the slope = mg/sin(alpha)

Since I can see that these 2 angles are indeed equal I am totally confused by these 2 different outcomes. 

What is wrong with my logic :/

m = 21 kg

α = 10°

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Yes!

$${\mathtt{FV}} = {\frac{{\mathtt{200}}{\mathtt{\,\times\,}}\left[{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{0.07}}}{{\mathtt{12}}}}\right)}^{\left({\mathtt{13}}{\mathtt{\,\times\,}}{\mathtt{12}}\right)}{\mathtt{\,-\,}}{\mathtt{1}}\right]}{\left({\frac{{\mathtt{0.07}}}{{\mathtt{12}}}}\right)}} \Rightarrow {\mathtt{FV}} = {\mathtt{50\,666.157\: \!719\: \!730\: \!719\: \!797\: \!1}}$$

You can just copy and paste the right-hand side of your expression into the calculator on the home page here.

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Thanks Mathcad, 

I have dug up this pic

okay - I think I have it   

1.  11011₂ = 1*2⁴ + 1*2³ + 0*2² + 1*2¹ + 1*2⁰ = 16 + 8 + 0 + 2 + 1 = 27

2.  999₁₆ = 9*16² + 9*16¹ + 9*16⁰ =2304 + 144 + 9 = 2457

3.  537₁₀ :  

16³ > 537

16² = 256, 2*256 = 512

537 - 512 = 25

25 = 16¹ + 9 or 16¹ + 9*16⁰

So 537₁₀ = 2*16² + 1*16¹ + 9*16⁰ = 219_h     (h for hexadecimal)

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$$log_2{(x+1)}=-3\rightarrow x+1=2^{-3}\rightarrow x=\frac{1}{8}-1\rightarrow x=-\frac{7}{8}$$

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7 over 8 times 5

well do it like this....times means multiplication and over means division!

7/8 x 5

now multiply as we cant reduce!

=35/8

i hope you can divide it yourself!

Okay, thanks!

There isn't a unique answer to this!

First, find how many times 6 (almost) goes into 80:  floor(80/6) = 13, and 13*6 = 78, so one solution is 13 goals and 2 behinds (13*6 + 2*1 = 80).

However, 12 goals gives 72 points so 12 goals and 8 behinds also gives 80 points.

11 goals gives 66 points so there would be more than 11 behinds needed to make 80 points in total, so this doesn't comply with more goals than behinds.

So the possible solutions are 13 goals and 2 behinds or 12 goals and 8 behinds.

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$${\frac{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{x}}\right)}}{\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{x}}\right)}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{x}}\right)}\right)}}$$      =   cot²(x)