All Answers 358108 Answers

11(2q-8)=-4(7-3q)   simplify

22q - 88  = -28 + 12q       add 88 to both sides, subtract 12q from both sides

10q =  60       divide both sides by 10

q = 6

(f + g)(2)  =  f(2)  + g(2)   .........notce that   f(2) means that we are just asking what "y" value is associated with x in the function "f" when x = 2  ......this is the point (2,3)......so  f(2)  = 3

Likewise, g(2)   = 1

So......  (f + g)(2) = f(2) + g(2)  =  3 +  1 =   4

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f(g(7)).....this means that, first, we are putting "7" into "g"  an finding the y value associated with that x value.....this is the point (7,4)....so   g(7) = 4

So....so far  ......f(g(7))  =  f(4).....now.......we are putting "4" into "f" and finding the y value associated with that x value.......this is the point (4,5)......so f(4)  = 5

To recap......f(g(7)  = f(4)  = 5

These are known as "composite" functions.....I'll admit they ARE a little tough to grasp !!!!

Post some more questions if you get stuck.....!!!

All you need to do to figure out the number, kid, is to add 100 + 38

which equals 138

$${\mathtt{100}}{\mathtt{\,\small\textbf+\,}}{\mathtt{38}} = {\mathtt{138}}$$

Subrtract 138 minus 100, and you get 38

$${\mathtt{138}}{\mathtt{\,-\,}}{\mathtt{100}} = {\mathtt{38}}$$

so, in order to figure out the number when it comes to adding and subtracting, do the opposite (+ transforms to -, vice versa.) this also applies to multiplication and division. 

The number you need to match the equation is 138

Here's the graph of the hyperbolic sine.........see for yourself where it = 0   ......

hi,

first thing you'd want to do is subtract 3 from both sides of your equation -

$${\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}} = {\mathtt{54}}$$

$${\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{x}} = {\mathtt{51}}$$

then the final step is to just divide 51 by 5!

$${\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{x}} = {\mathtt{51}} \Rightarrow {\mathtt{x}} = {\frac{{\mathtt{51}}}{{\mathtt{5}}}} \Rightarrow {\mathtt{x}} = {\mathtt{10.2}}$$

so your final answer would be 10.2!

hope this helps!

Heureka, I see what you did with that algorithm to find the relative primes to "n.'   I've not seen that done before........but....it's pretty neat.......is there a proof of this????

2sinpi/3cospi/3 ?

$$2\cdot\sin{ (\frac{\pi}{3}) }\cdot \cos{ (\frac{\pi}{3}) } = \sin{( ~2\cdot(\frac{\pi }{3})~ ) } =\frac{1}{2}\sqrt{3} = 0.86602540378$$

no tin this is my computer holiday homework......."holiday"homework!

2sinpi/3cospi/3=0

because sin pi = 0

Hallo Melody,

$$\small{\begin{array}{rcl}\mathbf{ \varphi(12)} &\mathbf{=}& \mathbf{4}$\\
\end{array}}$$,

because

$$\small{
\begin{array}{r|rcl|c}
\hline
\text{number} &\text{greatest common devisor} &&& \text{relative prime, if 1} \\
\hline
1 & \gcd{(12,1)} &=& \textcolor[rgb]{1,0,0}{1} & \text{yes}\\
2 & \gcd{(12,2)} &=& 2\\
3 & \gcd{(12,3)} &=& 3\\
4 & \gcd{(12,4)} &=& 4\\
5 & \gcd{(12,5)} &=& \textcolor[rgb]{1,0,0}{1} & \text{yes}\\
6 & \gcd{(12,6)} &=& 6\\
7 & \gcd{(12,7)} &=& \textcolor[rgb]{1,0,0}{1} & \text{yes}\\
8 & \gcd{(12,8)} &=& 4\\
9 & \gcd{(12,9)} &=& 3\\
10 & \gcd{(12,10)} &=& 2\\
11 & \gcd{(12,11)} &=& \textcolor[rgb]{1,0,0}{1} & \text{yes}\\
12 & \gcd{(12,12)} &=& 12\\
\hline
&&& \text{sum} &\textcolor[rgb]{1,0,0}{4}\\
\hline
\end{array}
}$$

Thanks Alan :)

Melody, hbar is Planck's constant h divided by 2pi and is usually written with the bar crossing the upright of the h rather than being above the h, thus:  $$\hbar$$

.

if m is doubled then

$$t=e^{-2aL}$$

would still be written the same way!  However, 'a' would be different.  So if we use 'a' for the original and 'a₂' for  when m is doubled then we could write:

$$t=e^{-2a_2L}$$

or

$$t=e^{-2\sqrt2aL}$$

.

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