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At first, Box F had 18 pears and 42 lemons while Box G had 36 pears and 50 lemons. Then, some lemons were moved from Box F to Box G and some pears were moved from Box G to Box F. In the end, Box F contained pears and lemons in the ratio 3:4 while Box G contained pears and lemons in the ratio 1:2

(a)  In the end, how many lemons were in Box F?

(b)  In the end, how many more pears did Box G contain than Box F?

Your "challenging question" is incomplete!!! 

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At first, Box F had 18 pears and 42 lemons while Box G had 36 pears and 50 lemons. Then, some lemons were moved from Box F to Box G and some pears were moved from Box G to Box F. In the end, Box F contained pears and lemons in the ratio 3:4 while Box G contained pears and lemons in the ratio 1:2.

(a)  In the end, how many lemons were in Box F?

(b)  In the end, how many more pears did Box G contain than Box F?

See this :

650 / 2 ==325

Sqrt(325)==~18 - this is the average of the 2 odd numbers.

18 - 1 ==17 - this is the smaller number.

18+1 ==19 - this is the larger number.

So that: 17^2 + 19^2 ==289 + 361 ==650

Let the number of worksheets printed on Tuesday==W
The number of worksheets printed on Monday==3/2W
The average for the following 3 days (Wed, Thu, Fri) was==[3/4 * 3/2W + 44]==9/8W + 44
3 * [9/8W + 44] + W==1287
27/8W + 132 + W ==1287
27/8W + W ==1287 - 132==1155
4.375W ==1155
W==1155 /4.375==264 - worksheets printed on Tuesday.
264 * 3/2 ==396 worksheets printed on Monday.

In triangle ABC, AB = 3 and BC = 15. Find the number of possible integer values of CA. (not including degenerate triangles)

Hello Guest!

\((15-3)<\overline {AC}<(15+3)\\ 12<\overline {AC}<18\)

\(\overline {AC}\in\{13,14,15,16,17\}\)

  !

-3 = 6a - 1 - 5a 

-3 = ( 6a-5a) - 1

-3 = a - 1

-3 +1  = a   -1 + 1

-2 = a

5 x 3 x 4     =  ............

(-3z^2-7z+4)(6z^2+z+6)

4 * z    +     6 * 7z           you can take it from here.....

Add the two equations together to get

4x + 4y + 4z = 1400    divied by 4

x + y  + z = 350        now divide by 3 to get the average of x , y and z............

I get      sqrt(405)

Do you still want/need an explanation?

A:B:C=3:2:3 (1)

so    C=A

and 

\(\frac{B}{A}=\frac{2}{3}\\ B=\frac{2}{3}*A\)

Now

15A+6B+9C=20 (2)

15A+6*(2/3)A+9A=20

15A+4A+9A=20

28A=20

\(A=\frac{20}{28}=\frac{5}{7}\\ B=\frac{2}{3}*\frac{5}{7}=\frac{10}{21}\\ C=\frac{5}{7}\)

The graph of y = f(x) is a parabola whose vertex is at (1, -2). The graph of y = f(x) - 3 is also a parabola. Where is the vertex of the graph of y = f(x) - 3?

Hello Guest!

\(y_1=f(x) =( x-{\color{BrickRed}1})^2{\color{BrickRed}-2}\ |\ \color{blue}vertex\ (1,-2)\\ \color{blue}y_1=f(x)=x^2-2x-1\)

\(y_2=f(x)-3 =( x-1)^2-2\color{blue}-3\)

\(y_2=f(x)-3 =( x-{\color{BrickRed}1})^2{\color{BrickRed} -5}\ |\ \color{blue}vertex\ (1,-5)\\ \color{blue}y_2=f(x)=x^2-2x-4\)

  !

When f(x) multiplies g(x) there will be a term 2x⁸.  No matter what size the constant b is, it won't affect the x⁸ term.  Hence the degree of the resulting polynomial will be 8.

Total = 9*a + 3*c

Now, just plug in 229 for a, and 385 for c.

4ax - 21x = -41x

(4a-21) x = - 41 x

4a - 21 = -41

4a = -20

a = -5 

A+B=17   (1)

B+C=16   (2)

(1)-(2)

A+B-B-C = 17-16

A - C = 1

so

A=1+C      

Now   

A*C = 30    so

(1+C)*C = 30

c^2+C-30=0

(C-5)(C+6)=0

C = 5 or -6

but C is positive so  C=5

so

A=1+C = 6

I think you can take it from there to find  B and D

Note that there is one piece of info in the question that I have not used yet.   

ahh tht answer is wrong thanks for ur help thoguh!

There were 25 people who went on the trip.

Expanding, we get that the coefficient of a is -3a + 28 = -41, so a = 23.

Attn: Helperid18...  and guest asker

Guest has asked the question as I have always requested that people do.

Thanks for that guest.

The only improvement I can see is that the address could be included as a link.

That way people could hyperlink to it more easily.

The asker should also request that people answer on the original and leave just a little note here (as well) stating that they have done so.

Thanks for answering Jugoslav :)

Thanks to gest for putting the link to Jugoslav's answer.

Hi Ginger.  It's wonderful to hear from you.  You keep track of my postings.  Since it's you, I'm flattered.  How do you know when it's me?  Do you recognize my style?  Or by the dot at the bottom?  I put the period down there, and color it white, just to add some white space at the bottom of the field.  It's been lonely not seeing any postings by you lately, or at least none in the easy problems which are all I'm capable of.  Ron 

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