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Hello. :D

This isn't as difficult as it first appears....although I have to admit, it threw me for a bit...!!!

Note that we have triangles ZAC and BAY

And two sides of ZAC - namely, b and c - are equal to two sides of BAY

And angle ZAB + angle CAB  = angle YAC + angle CAB

Thus....by SAS......triangle ZAC is congruent to triangle BAY....therefore ZC  = BY

Likewise...we have two other triangles, BCY and XCA

And two sides of BCY - namely, a and b - are equal to to sides of XCA

And angle XCB + angle ACB  = angle YCA + angle ACB

Thus.....by SAS......triangle BCY  is congruent to triangle XCA

Thus BY  = XA  = AX

But BY was also proved equal to ZC = CZ

Therefore....  AX = BY = CZ

P.S. - Melody should get some points for providing the diagram....without that, I wouldn't have understood the layout....!!!!

www.mathway.com is a wonderfull site!

Even though Michaelcai has the answer, it might be instructive to see what it is.....

Let the base of the triangle be AB  and let AD and BE intersect at right angles at F

 AF and FB  will be the legs right triangle AFB with AF = (2/3)AD and FB   = (2/3)(BE)

And AB will form the hypotenuse of AFB  and will  = sqrt [ (2/3)AD^2 + (2/3)BE^2 ] =

sqrt [ 10^2 + (40/3)^2 ] = 50/3

And  BD  will form a hypotenuse of triangle BFD  with BF = (2/3)BE  and FD  = (1/3)AD

And since D is a median on BC, BC  = 2BD =   2 sqrt [ (40/3)^2 + (5)^2 ] = 2 * 5 sqrt [ 73 ] / 3 =

sqrt (7300)/3

And AE will form a hypotenuse of triangle AFE  with AF  =  (2/3)AD and FE = (1/3)(BE)

And since E is a median  on  AC, AC = 2*sqrt [ 10^2 + (20/3)^2 ] = 2 * 10sqrt (13) / 3  =

sqrt(5200)/3

Now....since we know all the sides, we could use Heron's Formula to find the area, but all those roots would make this messy...!!!

Instead...lets use the Law of Cosines to find apex angle ACB....so we have that

AB^2  = BC^2 + AC^2  - 2 (BC)(AC)cos ACB

[50/3]^2  = 7300/9 + 5200/9  - 2 (sqrt(7300)*sqrt(5200)/9 * cos ACB

[ 2500 - 7300 - 5200] /9  =  -2 (sqrt(7300) *sqrt(5200) /9 * cos ACB

-10000  = -2 [ sqrt(7300)*sqrt (5200) ] * cosACB

arccos  [ 5000  /  [ sqrt(7300)*sqrt(5200] ] = ACB  ≈ 35.75°

So....the area of ABC  =  (1/2)(AC)(BC) sinACB  =

(1/2)sqrt(5200)/3 * sqrt (7300)/3 sin (35.75)   =  200 units^2

Here's a pic :

BTW....we can confirm that the answer is correct because C  = ( 4/3, 24)

So...the height of ABC  = 24 and the base  = 50/3

So.....the area  = (1/2) (24) (50/3)  = 12 * 50 / 3  = 600 / 3  = 200 units^2

FV = PV x [1 + R]^N

FV = 14,000 x [1 + 0.06/2]^(6*2)

FV = 14,000 x [1.03]^12

FV = 14,000 x     1.425760887.......

FV =$19,960.65

I rather like this question coming back up again.

I want to spend more time on questions like this.

Thanks again Heureka :)

Still Ginger has got a point, helperid, about you presenting you own answer or partial answer...

1/a + 1/b = 2/3.......................(1)
1/a + 1/c = 3/5....................... (2) 
1/a + 1/b + 1/c = 1..................(3) subbing 1 into 3 we get:
2/3 + 1/c = 1, and c =3 and subbing 2 into 3 we get:
3/5 + 1/b = 1, and b =5/2. adding the reciprocals of b, and c we have:
1/3 + 2/5 = 11/15, and the reciprocal of this is:
15/11 = 1 4/11 hours for B and C to finish the job.

We need someone better than me at geometry proofs here   

Any takers  ??

Thios is the diagram. 

The aim is to show that the 3 red lines are equal in length.

We need to find how many plus strokes (what are plus strokes?) he has in a minute. At least, I think you do... 😶

So use a ratio.

31 pulse strokes : 10 seconds = x pulse strokes : 60 seconds.

1860 = 10x

186 = x

Gustav's pulse is 186 pulse strokes per minute.

I didn't fully understand the question so I might be wrong.

Hi...?

You do just subtract the two.  The uppercase Greek letter \(\Delta\)  is often used to indicate the change in something; so \(\Delta d\)  means the change in d.

Well, I don't think this is an equation. An equation has an '=' sign. 

What you wrote is actually an expression.

Now you can't really solve this thing because there is no = sign.

But if the question told you to simplify the expression...

Distribute 3x and get:

3x*4b - 3x*2x.

Simplify to get 12bx-6x^2.

And that would be the answer if the question said "simplify the expression".

When printing multiples of 8, in what term will the 2017th "2" digit appear in?

In term 2906  (23248 = 2906*8)  will the 2017th "2" digit appear in

If A and B working together can finish a task in 1 1/2 hours,
and if A and C working together can finish the same task in 1 2/3 hours,
and if A, B, and C working together can finish the same task in 1 hour,
how long would it take for B and C working together to finish the same task?

Let W = Work

\(\begin{array}{|rcll|} \hline \frac{W}{A} &+& \frac{W}{B} && &=& \frac{W}{1\frac12\ h} \quad & | \quad : W \\ \frac{W}{A} & & &+& \frac{W}{C} &=& \frac{W}{1\frac23\ h} \quad & | \quad : W \\ \frac{W}{A} &+& \frac{W}{B} &+& \frac{W}{C} &=& \frac{W}{1\ h} \quad & | \quad : W \\ & & \frac{W}{B} &+& \frac{W}{C} &=& \frac{W}{x} \quad & | \quad : W \\\\ \frac{1}{A} &+& \frac{1}{B} && &=& \frac{1}{1+\frac12} & (1) \\ \frac{1}{A} & & &+& \frac{1}{C} &=& \frac{1}{1+\frac23} & (2) \\ \frac{1}{A} &+& \frac{1}{B} &+& \frac{1}{C} &=& \frac{1}{1} & (3) \\ & & \frac{1}{B} &+& \frac{1}{C} &=& \frac{1}{x} & (4) \\ \hline \end{array} \)

\(\frac{1}{A} =\ ?\)

\(\small{ \begin{array}{|lrcll|} \hline (1)+(2)-(3): \\\\ & (\frac{1}{A}+\frac{1}{B}) + (\frac{1}{A}+\frac{1}{C})-(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}) &=& \frac{1}{1+\frac12} + \frac{1}{1+\frac23} - \frac{1}{1} \\ & \frac{1}{A}+\frac{1}{B} + \frac{1}{A}+\frac{1}{C}-\frac{1}{A}-\frac{1}{B}-\frac{1}{C} &=& \frac{1}{1+\frac12} + \frac{1}{1+\frac23} - \frac{1}{1} \\ & \frac{1}{A} &=& \frac{1}{1+\frac12} + \frac{1}{1+\frac23} - \frac{1}{1} \\ & \frac{1}{A} &=& \frac{1}{\frac32} + \frac{1}{\frac53} - 1 \\ & \frac{1}{A} &=& \frac23 + \frac{3}{5} - 1 \\ & \frac{1}{A} &=& \frac{10+9}{15} - 1 \\ & \frac{1}{A} &=& \frac{19}{15} - \frac{15}{15} \\ & \mathbf{\frac{1}{A}} & \mathbf{=} & \mathbf{\frac{4}{15}} \\ \hline \end{array} }\)

\(x =\ ?\)

\(\begin{array}{|lrcll|} \hline (3) & \frac{1}{A} + \frac{1}{B} + \frac{1}{C} &=& \frac{1}{1} \\ & \frac{1}{B} + \frac{1}{C} &=& \frac{1}{1} - \frac{1}{A} \quad & | \quad \frac{1}{B} + \frac{1}{C} = \frac{1}{x} \\ & \frac{1}{x} &=& \frac{1}{1} - \frac{1}{A} \\ & \frac{1}{x} &=& \frac{1}{1} - \mathbf{\frac{4}{15}} \\ & \frac{1}{x} &=& 1 - \mathbf{\frac{4}{15}} \\ & \frac{1}{x} &=& \frac{15}{15} - \mathbf{\frac{4}{15}} \\ & \frac{1}{x} &=& \frac{15-4}{15} \\ & \frac{1}{x} &=& \frac{11}{15} \\ & x &=& \frac{15}{11} \\ & \mathbf{x} & \mathbf{=}& \mathbf{1\frac{4}{11}\ h} \\ \hline \end{array} \)

It would take for B and C working together to finish the same task \(\mathbf{1\frac{4}{11}\ h}\)

OR

\(log_{(4/9)}(\frac{27}{8})^{3/4}\\ =log_{(4/9)}(\frac{3}{2})^{9/4}\\ =log_{(4/9)}(\frac{2}{3})^{-9/4}\\ =log_{(4/9)}(\frac{4}{9})^{-9/8}\\ =\frac{-9}{8}\)

Let  the part of the job that A can do in one hour  = A

Let the part of the job that B can do in one hour  = B

And let the part of the job that C can do in one hour  = C

And using....... rate per hour * hrs. worked = part of the job done......

we  have this system of equations

[ Note....1.5 hrs = 3/2 hrs and 1 + 2/3 hrs  = 5/3 hrs  ]

(A) (3/2) + (B)(3/2)  = 1  →  3A + 3B  = 2   → B =  [2 - 3A] / 3    ( 1)

(A)(5/3) + (C)((5/3)  = 1  →  5A + 5C  = 3 → C =  [3 - 5A ] / 5     (2)

A + B + C  =  1               → A + B + C  = 1    (3)

Subbing (1) and (2)  into (3)...we have that

A + [2 - 3A]/3 + [3 - 5A]/5  = 1      multiply through by 15

15A  + 5[ 2 - 3A] + 3[ 3 - 5A] = 15    simplify

15A + 10 - 15A + 9 - 15A  = 15

-15A  + 19 = 15

-15A = -4    divide both sides by -15

A  = 4/15   ...so A can complete 4/15 of the job in one hour

B  completes  .... [2 - 3(4/15)] / 3  =  [ 30 -12] / 45  = 18/45  = 2/5  of the job in one hour

C  completes ...  [ 3 - 5(4/15)] / 5  = [ 45 - 20] / 75 = 25/75  =1/3 of the job in one hoiur

So....B and C working together complete    2/5 + 1/3  = [6 + 5] / 15  =

11/15  of the job in one hour

Flip this fraction over to find the total time it takes B + C working together to finish the whole job  = 

15/11 hrs ≈  81.81 minutes  ≈ 1 hr, 21minutes,48.6 seconds

Let

\(y=log_{(4/9)}(27/8)^{(3/4)}\\ y= \frac{log(27/8)^{(3/4)}}{log(4/9)}\\ y= \frac{(3/4)log(27/8)}{log(2/3)^2}\\ y= \frac{(3/4)log(3/2)^3}{log(2/3)^2}\\ y= \frac{(9/4)log(3/2)}{2log(2/3)}\\ y= \frac{9log(3/2)}{8log(2/3)}\\ y= \frac{9log(3/2)}{8log(3/2)^{-1}}\\ y= \frac{9log(3/2)}{-8log(3/2)}\\ y=\frac{-9}{8} \)

Invisable :))

That's a really good way of explaining it Gh0sty !! 

The layout of Heureka’s presentation is very clear. He enumerated the formula and its details to at least two prerequisite requirements.    

You’ve had two months to study this and said nothing until now; so, you should blôôdy well do it yourself and present it on here for evaluation, if you are unsure of your work. 

yes                                                                                                                         

Of course, if nothing else, you can always do \(x^\frac{1}{4}\)

No one will correct you here! You are 100% right. I did it myself.

I will solve it for you! The expression I will evaluate is \(\left(\frac{-2}{3}\right)^2\):

By the way, if you would like to type it into a calculator, do it like this: (-2/3)^2. The calculator will output 0.444...