All Answers 358514 Answers

PLZ, HELPPPP!!!!!!

Hi MG, 

It is not rude to ask for explanation.        

I am very pleased that you are learning from me.  That is what I am here for.  -   to teach people 

Plus intelligent, enthusiastic students like you are fun and rewarding to teach. 

 $$y=8\times \left(\frac{2}{4}-\frac{\frac{6^3}{2}}{15}\right)$$

\frac{\frac{6^3}{2}}{15}

$$\frac{\frac{6^3}{2}}{15}$$

I have 'nested' a fraction inside another fraction,  that is why there are two \frac   

The black parentheses belongs to the main fraction.  The red parentheses belong to the nested fraction.

21?     

$${\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}} = {\mathtt{2}}$$

Cool but a little dumb...

LOL!

$${\mathtt{2}}{\mathtt{\,-\,}}{\mathtt{1}} = {\mathtt{1}}$$

I can't find an exact formula for the sum of N terms, but there is an approximate expression:

sum ≈ 2*N - 1.02  which gives it to two decimal places.

or sum ≈ 2*N - 1.020004 which gives it to 6 decimal places for N>3

.

Great answer Melody, I didn't even notice that box.

Sweet grid work!!!

I assume you mean it's inscribed in a circle.  

If so, chop it up into isosceles triangles, where the apex of each triangle is at the centre of the circle, and the two equal sides of each triangle are radii, length r, of the circle.

The base of a triangle with apex angle, A, is given by 2r*sin(A/2).  

If it's a regular polygon, with n sides, then A = 360°/n.

.

Melody, is obviously Melody. 

She always uses perfect English.

" I don't know how that person hacked me...now I had to change my account and everything. That's why there are two Melodys. The other one is an imposter. Stranger Danger! "

Where is the caps after ... now?

:D

y=8\times \left(\frac{2}{4}-\frac{\frac{6^3}{2}}{15}\right)

I have some questions to do with this code:

• Near the 2 (after 6^3) why is there two ]} ?
• After the - (after {2}{4}) why is there two frac sign?

Sorry if it sounds rude, but I just want to know the reason why.

Thanks :)

Using the factv that log₁₀(x) =  a means x = 10, and here we have a = -8.3 then

x = 10^-8.3 ≈ 5*10^-9

.

Since the only force acting on the ball after it has been kicked is gravity we can use the constant acceleration equations:

v² = u² - 2*g*s   ...(1)    where v = final velocity, u = initial velocity, g = 32 ft/sec² and s = distance travelled

and

v = u - g*t      ...(2)     where t = time

Maximum height is when v = 0, so, using (1):

0 = 63² - 2*32*s

s = 63²/(2*32) ≈ 62 ft

But the ball was kicked from astarting height of 2 ft, so maximum height is 64 ft.

Using (2) we have 

0 = 63 - 32*t

t = 63/32 ≈ 1.97 seconds

An athletics track is to have a perimeter of 400m and straights of length 100m. Find the radius of the bend ?

$$\small{\text{$
\begin{array}{rcll}
2\cdot (\pi \cdot r + 100~ \rm{m} ) &=& 400 ~ \rm{m} \quad & | \quad :2 \\
\pi \cdot r + 100~ \rm{m} &=& 200~ \rm{m} \\
\pi \cdot r &=& 100~ \rm{m} \\
r &=& \dfrac { 100~ \rm{m} } { \pi } \\
r &=& 31.8309886184 ~ \rm{m} \\
r &\approx& 31.83 ~ \rm{m}
\end{array}
$}}$$

Thank you CPhill.

The same goes for you everyday.

Also, did you have a large increase in points due to the system update?

It seems like only i did because my points was close to be doubled (560-1060)

Amount = Principle*(1 + rate)^Times compunded 

A=P(1+r)^n

For 5 years:

A = 4500*(1+0.065)^5

A = $6165.39

For 12 years:

A = 4500*(1+0.065)^12

A = $9580.93

You've left out the k!  Where does it go?

.

$15,334.79 ( 1 + .065 / 365)^600  = about $17063.88  

Looking at my series answer.......the distances narrow.....thus....he was squashed between the handlebars....

Alas.....all that flying was for naught.....!!!!!

A better question is

Where was the fly when the cyclists met ? 

divide and write the result using scientific notation (4.0 x 10^7) / (5.0 x 10^9)

$$\small{\text{
$
\begin{array}{rcl}
\dfrac{4.0 \cdot 10^7} {5.0 \cdot 10^9} & \\
&=& \dfrac{4.0}{5.0} \cdot 10^7\cdot 10^{-9} \\
&=& 0.8 \cdot 10^{7-9} \\
&=& 0.8 \cdot 10^{-2} \\
&=& \dfrac{8}{10} \cdot 10^{-2} \\
&=& 8 \cdot \dfrac{ 10^{-2}}{10} \\
&=& 8 \cdot 10^{-2}\cdot 10^{-1} \\
&=& 8 \cdot 10^{-2-1} \\
&=& 8 \cdot 10^{-3 }
\end{array}
$}}$$

The answer is one.

The answer to anything to the power of zero is one.