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"The odometer in my car is currently 84095.
It advances by ones.
The tripometer is at 8018.6.
It advances by tenths.

When will the numbers look "the same"?
For example 84523 and 8452.3

And what is the algorithm to solve this?"

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You are right not to be keen on Euler's method, especially with a step size this large - see the comparison below:

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What is the equation of the parabola passing through (1,5), (0,6), and (2,3)?

Formula parabola:

\(\begin{array}{|rcll|} \hline y = ax^2+bx+c \\ \hline \end{array} \)

a, b, c = ?

\(\begin{array}{|lrcll|} \hline P(0,6): & 6 &=& 0^2\cdot a+0\cdot b+c \\ & 6 &=& c \\\\ P(1,5): & 5 &=& 1^2\cdot a+1\cdot b+c \\ & 5 &=& a+b+c \quad & | \quad c=6 \\ (1) & 5 &=& a+b+6 \\\\ P(2,3): & 3 &=& 2^2\cdot a+2\cdot b+c \\ & 3 &=& 4a+2b+c \quad & | \quad c=6 \\ (2) & 3 &=& 4a+2b+6 \\ \hline \end{array} \)

a, b = ?

\(\begin{array}{|rcll|} \hline (2) & 3 &=& 4a+2b+6 \quad & | \quad : 2\\ & 1.5 &=& 2a+b+3 \\\\ (1) & 5 &=& a+b+6 \\ \hline (2)-(1): & 1.5-5 &=& 2a+b+3- (a+b+6) \\ & -3.5 &=& 2a+b+3- a-b-6 \\ & -3.5 &=& a-3 \\ & -0.5 &=& a \\\\ & 5 &=& a+b+6 \quad & | \quad a=-0.5 \\ & 5 &=& -0.5+b+6 \\ & 5 &=& 5.5+b \\ & 5-5.5 &=& b \\ & -0.5 &=& b \\ \hline \end{array} \)

Formula parabola:

\(\begin{array}{|rcll|} \hline y &=& ax^2+bx+c \quad & | \quad a=-0.5 \quad b=-0.5 \quad c=6 \\ \mathbf{y} & \mathbf{=} & \mathbf{-0.5x^2-0.5x+6} \\ \hline \end{array}\)

As follows:

So f(2) = y(2) = 6

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There is also a generalization of this question:

aⁿ+bⁿ/(ab)^n-1+1 is an integer k. Prove that k=cⁿ where c is also an integer (a, b and n are all integers)

Hey, I'm studying for a test on Wednesday.

Would I be able to cancel out the "^16" in this equation to get lim_(x->∞) (x + 1)^16/x^16 = 1?

\(\begin{array}{|rcll|} \hline && \lim \limits_{x\to \infty} \dfrac{ (x + 1)^{16} } {x^{16}} \\ &=& \lim \limits_{x\to \infty} \dfrac{ \binom{16}{0} x^{16} + \binom{16}{1} x^{15} + \binom{16}{2} x^{14} + \ldots + \binom{16}{15} x^1 + \binom{16}{16} } {x^{16}} \\ &=& \lim \limits_{x\to \infty} \dfrac{ 1\cdot x^{16} + 16 \cdot x^{15} + 120 \cdot x^{14} + \ldots + 16\cdot x + 1 } {x^{16}} \\ &=& \lim \limits_{x\to \infty} \frac{ x^{16} } {x^{16}} + 16 \cdot \frac{ \cdot x^{15}} {x^{16}} + 120 \cdot \frac{ x^{14}} {x^{16}} + \ldots + 16\cdot \frac{ x } {x^{16}} + \frac{ 1 } {x^{16}} \\ &=& \lim \limits_{x\to \infty} 1 + \frac{ 16} {x} + \frac{ 120} {x^2} + \ldots + \frac{ 16 } {x^{15}} + \frac{ 1 } {x^{16}} \\ &=& 1 + 0 + 0 + \ldots + 0 + 0 \\ &=& 1 \\ \hline \end{array} \)

Oh yeah! I see that now too..

I really like the circle solution though.. it's very clever! 

Well, I was trying to find a " real " formula for it, by searching for " function for the duration of a storm " in Google, but all I actually found was several others asking the same question on different sites !

Thanks, hectictar.......where did you find that "formula??".....I think it would be interesting to find out how that was derived....!!!!

Not to beat a dead horse, but.....here's a way easier solution based on hectictar's pic.....no circle drawing is actually required!!!!

Note that triangle CBG is isosceles with angle  CBG  = 150

Since CB  = BG.....so...  angle BCG  =  [ 180 - 150] / 2   =  15

But angle BCE  =  60  =   angle BCG + angle GCE

So

60  = 15  + angle GCE  

45  =  angle GCE 

I think heueka may know how to do this......WolframAlpha does not give a nice, neat "exact" value.....but......that doesn't mean that one doesn't exist.....!!!!!

I think this is the function, since I looked it up:

\(T(D) = 0.07D^{\frac32}\)

where D is the diameter of the storm in miles and T is the time in hours.

To find the duration of a storm with a diameter of 1 mile,

just plug in 1 for D and solve for T(1).

\(T(1) = 0.07(1^{\frac32}) \\~\\ T(1) = 0.07(\sqrt{1^3}) \\~\\ T(1)=0.07(\sqrt1) \\~\\ T(1) = 0.07 \text{ hours}\)

Thanks for that pic, hectictar........your illustrations bring "life" to these answers....!!!

Wow, that's really good!

I would not have thought to draw a circle ! 

Here is a picture, I thought I might as well share it:

v(6.9)   =  -102.357

v(1.9)  = -36.8025      so

[-102.357  -  (-36.8025) ]  /  5    = -13.1109  

you can write this into an equation

# of 30 second ads= x

# of 60 second ads= y

x+y=13

x175,000+y250,000 = 2,650,000
 

we can turn x+y=13 into x=13-y

and replace x in the second one with such

giving us y250,000 - y175,000 + 2,275,000 = 2,650,000

by subtracting 2,275,000 from both sides and simplifying the left side we are given

y75,000 = 375,000

divide both sides by 75,000

y=5

now we plug that into x+y=13

x+5=13

subtract 5 from both sides

x=8

which means 8 30 second ads, and 5 60 second ads.

This question is not fully asked, mind extending?

Draw a circle with B  as a center and BA as a radius....

This circle will pass through the points  A, C, E  and G

And angle EBG  will  be a central angle that will intercept the minor arc  EG

And the measure of this angle = 90°

And angle  GCE  will be an inscribed angle  intercepting the same arc as  angle EBG 

And an inscribed angle intercepting the same arc as a central angle will have 1/2 the measure of that angle......so.....  angle GCE  =  (1/2) (90°)  =  45°

Thanks Ginger :)

It is great being taken to old posts like these.

I love history LOL  :))

I am sorry, I do not know how to find the exact value of cos80º - cos25º

But I am interested in this problem..if no one else knows how to do it, can you ask your teacher about it?

If there is a way to find the exact value of it I would be very interested to know it!!! 

\(145\frac56\,*\,6\)

Convert the mixed number to an improper fraction.

\(145\frac56\,*\,6=\frac{875}{6}\,*\,6\)

If you take some number and divide by 6 then turn around and multiply by 6, you end up with the exact same number you started with.

\(145\frac56\,*\,6=\frac{875}{6}\,*\,6=875\)

Maybe I can put this picture to good use:

Now we can see that

θ + 58º = 90º

θ = 90º - 58º

θ = 32º

I don't really know what you mean by "for the acute angle in a right triangle," since both of the other angles in a right triangle are acute. But there are 2 angles that have the same sin as sin(32º) .

Those two angles are:

32º  (acute)      .....and.....      180º - 32º = 148º (obtuse)    

The slope of a line that passes through (26 , 1052) and (-53 , -1950) = \(\frac{\text{change in y's}}{\text{change in x's}}=\frac{1052-(-1950)}{26-(-53)}=\frac{3002}{79}=38\)

The equation of this line is

(y - 1052) = 38(x - 26)

y = 38x - 988 + 1052

y = 38x + 64

Now, just plug in -92 for x and solve for y.

y = 38(-92) + 64 = -3496 + 64 = -3432

So, the point with an x-coordinate of -92 is : (-92 , -3432)

Yes you are right, I have written it back to front but my logic and my answer is still correct.

I shall attempt to fix my wording :)