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So our equation was right, we just couldn't figure out how to solve it.

g ( f(x) ) =   2 [ ax + b ] - 5   = 2ax + 2b - 5

So....we have

2ax + 2b -  5  = 3x + 4

This implies that

2ax = 3x  ⇒   2a = 3 ⇒  a = 3/2

And

2b - 5 = 4    add 5 to both sides

2b = 9       divide both sides by 2

b = 9/2

So

a + b   =     3/2 + 9/2    =  12 / 2    =   6

Hmm, if you match both equations... Let's see what CPhill says.

it turns out that if I take some number, call it x, and that number is between 0 and 1, then

\(1 + x + x^2 + x^3 + \dots = \dfrac{1}{1-x}\)

if as in this case x=1/2 we have

\(1 + \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 8 + \dots = \dfrac{1}{1-\dfrac 1 2}= 2\)

and subtracting 1 from both sides

\(\dfrac 1 2 + \dfrac 1 4 + \dfrac 1 8 + \dots = 1\)

Now you know some math I bet your friends don't!

are you allowed to just do that, though? Just put in a random value?

Wait, how about if you equal 2b=4, so b=2. It will be better if you match both sides! Also, 2ax=3x, x will cancel, thus 2a=3, a=3/2. Therefore, 3/2+2=7/2.

Some of the moderators could probably solve this.

I don't know, it's probably some complicated stuff that needs some sort of manipulation. I'm only a 6th grader doing 7th grade level math, and one who will be doing 9th grade level math next semester, okay?

Using the Law of Sines

sin B            sin C

____    =    _____

 AC             AB

sin 145           sin C

______   =    _______

2035             1017.15

sin C   =  (1017.5) * sin (145) / 2035

arcsin   [ (1017.5) * sin (145) / 2035 ] =  C = 16.67° ≈ [1.67 x 10^1 ] ° 

YAY! I got it! Phew, I did it! Thank you!

That's what I did! How do you proceed from there, though?

I have no idea, but try this:

2(ax+b)-5 is supposed to equal 3x+4, right? From there, it is just algebra.

(don't send hate, I'm 11)

It is 5/7(as seen below)

I found it:

The only time the two numbers selected will not have a positive difference which is 2 or greater is when the two numbers are consecutive. There are 6 pairs of consecutive numbers in the set, and there are (\($\binom{7}{2}=21$\) pairs of numbers total. So, the probability that the pair of numbers chosen is not consecutive is  1-6/21=5/7.

is it 5/7?

Try again; you're close through. 

This answer is incorrect.

There are \(7C2=\frac{7!}{5!2!}=21\)  possibilities of number pairs.

For the difference to be 2 or greater, the numbers cannot be 1 apart. Therefore, the sets {1, 2}, {2, 3}, {3, 4}, etc. all the way to {6, 7} cannot be used. Counting, we have 6 pairs that don't work so \(21-6=15\) and the probability is \(\frac{15}{21}=\frac{5}{7}\).

You are very welcome!

:P

I'm 11 and don't understand any of that, Rom.

Zeno missed a chance to make a profound (for the time) discovery about infinite series.

\(\text{let the width of the street be }L\\ \text{via Zeno's semi correct reasoning }\\ L = \dfrac{L}{2} + \dfrac{L}{4} + \dots = L \sum \limits_{k=1}^\infty~\left(\dfrac 1 2\right)^k \\ \text{i.e. } \sum \limits_{k=1}^\infty \left(\dfrac 1 2\right)^k = 1\)

Thanks! ;)

Hmmm... That's weird! Close I guess!

From the information that you have given, it is angle A=16.67 degrees, NOT angle C!.
 Calculation of the inner angles of the triangle using  Law of Cosines
A = b^2+c^2 - 2bc cos( A ) 
A = arccos(  a^2-b^2-c^2 }{ 2bc } ) = arccos(  1017.5^2-2035^2-1116.03^2 }{ 2 * 2035 * 1116.03 } ) = 16° 39'57" 
                                                                                                              Sides: a = 1,017.5 b = 2,035 c = 1,116.031

Area: T = 325,665.555
Perimeter: p = 4,168.531
Semiperimeter: s = 2,084.265

Angle  A = 16.666° = 16°39'57″ 

Angle  B = 145°                                             
Angle  C = 18.334° = 18°20'3″ 

I just typed your answer into a calculator and found that your answer was actually 9.99999-something!

D of course.

And the regina profile pic is awesome by the way.