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We have the following equation

( 36B + 6B + A )  + ( 4*36 + 1* 6 + B)  + (36A  + 6  + 5)  = 216A + 36 + 5*6 + 2

Simplify

43B  + 37A  + 150 +  11  = 216A + 68

43B + 37A  + 161  = 216A + 68

43B   = 179A - 93

Note....this equation will be true when B  = 2  and A  = 1

So

l A  - B  l  =   l 1   - 2   l    =  l  -1  l   =   1

I'm not seeing any simple way of doing this though the answer is simple enough it makes me suspect there is one.

(a)  We have the points  on the graph  (0, 0 )  and ( 3,6)

An equation for this  is  C = (6/3)x  = 2x

(b) We have the points on the graph (0,0)  and (4.6)

An eqution for this is  C = (6/4)x  = (3/2)x

(c)  Snce the"strawberry" graph  is "above"  the "blueberry" graph at  every point  [except at (0,0)] ....the blueberries are cheaper at any comparable weight to the strawberries

We have  two triangles to consider

First  ...triangle  AGF

Since JE  is a segment drawn parallel to base AG.....we have the following relationship :

FE / JE   = FA / AG

1 / JE  = FA / AG

1/ JE  = 5 /AG

AG = 5  (JE )        (1)

Next...we have triangle AGD

Since  HC  is a segment drawn parallel to  base AG...we have the following relationship :

DC / HC  = DA / AG

1 / HC  =  3 / AG 

AG = 3 ( HC )       (2)

Equating (2) and (3)

5 ( JE )  = 3 ( HC )

HC / JE  = 5 / 3

NVM I GOT IT!

Here's the one I like  :

https://matrix.reshish.com/gauss-jordanElimination.php

You need to know the basic equation of ballistic motion

\(h(t) = -\dfrac{g t^2}{2} + v_0 t + h_0\)

here they give you \(h_0 = 60\)

they also seem to be using \(g = -10 ~m/s^2\) which isn't quite accurate but close enough for government work.

You have two equations

\(h(2)= 80,~h(6) = 0\\ \text{you really only need one of them, plugging in values we have}\\ 80 = -5(2^2) + 2v_0 + 60\\ 20 = 2v_0 -20\\ v_0 = 20 \\ \text{checking with the other equation we have}\\ h(6) = -5(6^2) + 20(6) + 60 = \\ -180 + 120 + 60 = 0 \text{ as expected. So our equation of motion is}\\ h(t) = -5t^2 +20t + 60 \\ \text{we can rewrite this as}\\ h(t) = -5(t^2 - 4t - 12) = \\ -5((t-2)^2-4-12)=\\ -5(t-2)^2 + 80 \\ \text{and this last is the standard form of the parabola }\)

Find the  proportionality constant,  k

y  = k / √x

4  = k / √2

4√2  = k   = √32

So

1 =  √32 / √x

√x  = √32

x  = 32

x^3 + 5x^2 - 9x  - 45

We can factor this by grouping

x^2 (x + 5)  - 9 ( x + 5)       [ ( x + 5) is the common factor] 

(x + 5) ( x^2 - 9 )

(x + 5) ( x + 3) ( x - 3)

So....only  ( x + 3)  is a factor

Δ ABC    ~   Δ ACD

So

AB /AC  = AC/ AD

AB / 6 = 6 / 3     multiply both sides by 6

AB  =  36 / 3   =   12

Id say 0 but I would need the rest of the question to answer properly.

If the expression (x-a) is a factor of f(x), then f(a) = 0.

So just plug each of them into f(x)

f(5) = 160

f(-3) = 0

f(15) = 4320

\(\dfrac{y_2}{y_1}=\dfrac{1}{\sqrt{\dfrac{x_2}{x_1}}}\\y_2 = \dfrac{1}{\sqrt{\dfrac{x_2}{x_1}}}y_1 \\ 1 = \dfrac{1}{\sqrt{\dfrac x 2}} 4 \\ 1 = 4 \sqrt{\dfrac 2 x}\\ 1 = 16 \dfrac 2 x \\ x = 32 \)

Thank you everybody, this is my thought process

\(BBA_6+41B_6+A15_6=A152_6 \\ \text{looking at the 1's digit we have, (all numbers will be base 6)} \\ A+B+5 =2 \mod 6 \\ \text{given the possible ranges of A and B we have}\\ A+B = 9 \text{ with a carry of 2 or }\\ A+B = 3 \text{ with a carry of 1}\)

\(\text{let's suppose the first case}\\ \text{then looking at the 6's digit we have}\\ B+1+1+2 = 5 \mod 6\\ B+4 = 5 \mod 6 \\ B = 7 \text{ or }B=1 \\ \text{well B=7 is clearly out as B<6, and A+B=9, which would make A=8, which is also out}\)

\(\text{so let's assume the 2nd case}\\ \text{looking at the 6's digit we have}\\ B+1+1+1 = 5 \mod 6\\ B+3 = 5 \mod 6\\ B=8 \text{ or }B=2 \\ \text{again B=8 is clearly out. If B=2 then A=1 and this seems like a solution}\)

\(\text{checking we have }\\ 221 + 412+115 = 1152 \text{ all in base 6}\)

\(|A-B| = |1-2|=1\)

I agree with you. Identifying the similar triangles is the toughest step. I had some trouble myself. 

Anyway, we know  \(\overline{AG}\parallel\overline{CH}\) from the given info. This means that corresponding angles are congruent. Therefore, \(\angle A\cong\angle HCD\) . Via the reflexive property of congruence, \(\angle HDA\cong\angle HDA\) . Can you think of two triangles that would be similar knowing this? We are hunting for two triangles that share a vertex at \(\angle HDA\) . This is the part that I initially overlooked. I hope this hint hel

Hint: Find the interior angle in a decagon.

Thought process: Using the formula, 180(n-2), 180(8)=1440-total degrees. We can divide by 10 to find the interior angle of each side, 1440/10=144. 

And, if you see the line is cutting exactly in the middle, so we have to x angles. We also have 2 decagon angles, which sum to 144(2)=288.

Since there are 360 degrees in a quadrilateral, we yield with the equation of 2x+288=360, 2x=72, x=36 degrees.

I hope I didn't make any silly mistakes! 

Why do you think it is 36?

Hi shteyas1,

I would have used a few more words like

And why exactly are those two triangles similar ?

BUT your logic is good.

I may be wrong, but is the answer 36. If it's correct, just tell me.

That is the part where I am stuck on because finding the similar triangles is so hard

Probably the toughest thing to do here is to find a pair of similar triangles that will be of benefit to you for this particular problem. Try your best to locate similar triangles. Use the given info about the parallel lines to help you. 

Here we go

Yes it is thank you for proving my answer to be right, this is my thought process

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