All Answers 358514 Answers

Oh, btw, its only for the tributes

no anons may answer it unless contestants get it wrong, and nobody thats not in contest cant do this. And plz do the math and send it to me via private message.

Doing it in public will result in immediate elimination.

Sorry if i made it late

i had to wait till all contestants(tributes) are offline

anyways, heres the first problem, which you all need to know

everyone that does this problem correctly within 24 hours gets a point from me (plz do it via private message)

Imagine yourself going to chevron. You fill 6.090 gallons in your car/truck and end up paying $18.26. Find the cost per gallon.

I assume you want to know how to access the tangent inverse function.....just hit the "2nd" key and you will see the "atan" function key ......  (tangent inverse key)

Let's call the time it takes the first car to travel the dstance, t.

Then, the time the second car travels is (t - 3.7) hrs.    ..... and since the distances are the same, we have

54(t - 3.7)  = 41t      simplify

54t - 199.8  =  41t     rearrange

54t - 41t = 199.8      simplify

13t = 199.8        divde both sides by 13

t =  about 15.369 hrs

$${\mathtt{10}}{\mathtt{\,\times\,}}{{\mathtt{8}}}^{-{\mathtt{5}}}$$

For whatever the number, divide it by Pi to get how many times bigger than Pi it is (Hence giving as a multiple of Pi).

Notice that the sum of the first n terms and the next term to be added are relatively prime, since their difference is just 1.  For instance:    1  + 2  + 4  = 7.  And the next term to be added is 8.

But, if A, B are relatively prime, then their sum, A + B, (the next  term in the series), must be relatively prime to B, because their difference is just A. [If A and B weren't relatively prime, there would be some factor, k, that would divide them both and would also divide their sum.]

Thus, each successive term in the series is relatively prime to its successor........

tan(x) = 11/r

cos(x) = r/s

sin(x) = ?

Using the equation:

$$Tan(x) = Sin(x) / Cos(x)$$

Therefore:

11/r = Sin(x) / (r/s)

11/r = (s * Sin(x)) / r

Since both sides are divided by "r", let's multiply it out.

11 = s * Sin(x)

We want Sin(x) on it's own, so divide both sides by "s".

11 / s = Sin(x)

Sin(x) = 11 / s

41_mph * t_hours = Total distance

54_mph * 3.7_hours = Total distance

Therefore: 41_mph * t_hours = 54_mph * 3.7_hours

Divide both sides by 41 to make "t" the subject of the equation.

t = (54 * 3.7) / 41

= 199.8 / 41

= 4.8731707317073171

= 4.9_hours (to 1 d.p.)

In the calculator  calculator you just type sqrt(

Ok.                                              

i dont know about "it" but this was invented.

Hello anonymous!

(2x^2+3x-1)+(-2x^2-x-2)

= 2x² + 3x - 1 -2x² - x - 2

= 2x - 3

(2x^2+3x-1)+(-2x^2-x-2) = 2x - 3

regards  :- )

Nit-picker !

But you're correct, I was careless, I should have stated that k was positive.

Here's another way to do this based on Alan's drawing...

The hypotenuse of the the small triangle at the bottom left = √[8^2 + 8^2] = 8√2

So, the length of the hypotenuse of larger triangle at the top right is just 27 - 8√2

And using similar triangles, the length of each side of this larger triangle = w =  [27 - 8√2]/ √2  = about 11.092 ft.  ≈  11.1 ft

33)    0.25/2ft = x/15ft    where x is the inches on the scale drawing   multiply by 15 on both sides

(15)(0.25)/ 2 = x  = 1.875 in

34)   The tiled area  = Total area - area occupied by cabinets.......

The total area is 15 x 12 =  180 sq ft

The area occupied by the cabinents is (12 x 2)  + (4 x 2 x 2)  =   24 + 16 = 40 sq ft

So  ...   180 - 40   =  140 sq ft  will be tiled (not covered by cabinets)

35) I'm taking a guess here as to what is meant......

10 cabinets will be installed.....so......the cost per cabinet is  $(2150 - 650)/ 10   =  $150  per cabinet

So  20 cabinets plus the fixed labor charge of $650  =  20 x $150 + $650  = $3650

First of all thank you so much for making thi big confusing line it I a small simple n nice easy definition.

secodly , what do u think? Am I a 2nd grade kid that I wouldn't know what MS Word and and app would be. Ah even 1st graders know what they are! Anyways forget this , I ha asked for the simplest definition I think you were going with my words.

now for this I will have to design the opening screen or the face of an application!now for me something hard to decide is hat application! I have many apps but which one. And how will I get its contents? What fo you suggehst? Which applications look should I destiny in MS Word? 

$${{\mathtt{x}}}^{{\mathtt{2}}} = {\mathtt{400}}$$

$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{400}} = {\mathtt{0}}$$

$$\left(\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{20}}\right){\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{20}}\right)\right) = {\mathtt{0}}$$

$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{x}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{x}}={\mathtt{20}}\\
{\mathtt{x}}=-{\mathtt{20}}\end{array}\right)}$$

x-2y=1                        x^2+y^2=13

                                   x ^2 = 13-y^2

                                    x  = sqrt(13y-2)

sqrt(13y-2) = 1+2y    | quadrieren

13-y^2 = 1+4y+4y^2

5y^2+4y-12=0

y1= 1,2          y2= -2

x=1+2y         x1 = 3,4       x2 = -3

$$\\x-2y=1\qquad (1a)\qquad x^2+y^2=13\qquad (2)\\
x=1+2y\\
x^2=1+4y^2+4y\qquad (1b)\\
sub\;1b\;into\;2\\
1+4y^2+4y+y^2=13\\
5y^2+4y-12=0\\
5y^2+10y-6y-12=0\\
5y(y+2)-6(y+2)=0\\
(5y-6)(y+2)=0\\

y=+6/5\;\;\;\;or\;\;\;\;y=-2$$

now sub thes y values back into 1a to find the corresponding x values.

you can finish it :)

Thanks Bertie, that is really neat.  It is so easy when YOU do it.   

s,k and t must all be positive integers I think.   :/

This is given by acceleration =  v²/r  where v is speed and r is radius.

I'll let you plug in the numbers!

.

It's a geometric series with first term 1 and common ratio 2, so

$$S_{n}=2^{n-1}-1,$$

and

$$S_{n+1}=2^{n}-1.$$

Suppose that both of these are exactly divisible by some integer k (say), then there will be integers say s and t such that

$$s\times k = 2^{n-1}-1, \; \text{ and }\; t\times k = 2^{n}-1.$$

Multiply the first of these equations by 2 and subtract that from the second one and you have

$$(t\times k)-(2\times s\times k) = 1,$$

or,

$$k(t-2s)=1.$$

Since k, t and s are integers, it follows that

$$k = 1 \; (\text{and }\; t - 2s=1).$$

@@ End of Day Wrap  Thurs 25/6/15   Sydney, Australia Time   10:45pm  ♪ ♫

Hi everyone,

Our answer tributes for the past two days are bestowed on: Heureka, MathsGod1, Radix, Sir-Emo-Chappington, Nauseated, DarkBlaze347, Civoamzuk, Alan Anonymous4338, Hades, AydenSchnabel, TitaniumRome, Fiora and asinus.   Thank you  

Sun 28/6/15

If you would like to comment on other site issues please do so on the Lantern Thread.  Thank you.    

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