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On 8X calcs, X delaults to 0; no vars, either.

1. Two cars start from the same point traveling in opposite directions. One travels 12 mph slower than the other. After 3 hours, they are 300 miles apart. Find the rate of each car.

2. Mr. Still spends 1 and 2/3 hours traveling to and from work. His average rate going to work is 45 mph and his average rate returning from work is 30 mph. How far does Mr. Still travel to work?

3. A jet flew 3,000 km in 6 hours. It flew for 2 hours before encountering a head wind. The head wind reduced the rate of the jet by 60 km/h. Find the rate of the jet before it encountered the head wind.

4. A boat can average 10 mph in still water. It takes twice as long to go 45 miles upstream as it does to go the same distance downstream. Find the rate of the current.

5. A car traveling 50 mph went 40 miles farther than a car traveling at 60 mph. The slower car traveled 3 hours longer than the faster car. Find the time each car traveled.

1) One car travelled the same distance of the other car, but travelled 36 miles further (distance = speed x time) so 264/2 miles is the total distance travelled by the slower car in 3 hours = 132 and the speed of that car was 44 mph. Other car was 12 mph faster so it was 56 mph.

2) This one I just kinda did in my head without calculations the numbers were quite clean to work with, and I matched the 1 hour time to the 30mph and the 2/3 hour time to the 45 mph journey and if he travelled at 30mph for 1 hour then the distance is 30 miles.

5) 50 = (D + 40)/(T + 3) (1)
60 = D/T (2)

(1) x (T + 3)
=> 50T + 150 = D + 40
=> 50T = D - 110 (3)

(2) x T
=> 60T = D (4)

(4) - (3)
=> 10T = 110

=> T = 11 (Time of the faster car)

11 + 3 = 14 (Time of slower car)

Only ones I had time to do without getting help from textbooks and the like

You are awesome.

But... I have to show (excrutiatingly detailed) work for all of them... I managed to figure it out for the first one, but I don't understand how you got the second one, and the last one... I have no idea what you did there. I know, I'm extremely slow, but... eh. Thank you, though!

2. Let miles be the distance required.

x/45 + x/30 = 5/3
(2x+3x)/90 = 5/3
5x/90 = 5/3
15x = 450
x=30

3. Let the original rate be y km/h.

2y + 4(y - 60) = 3000
2y + 4y - 240 = 3000
6y = 3240
y = 540

4. Let r mph be the rate of the current.

2 * 45 (10 - r) = 45 (10 + r)
20 - 2r = 10 + r
10 = 3r
r = 10/3

5. Let the time the faster car travelled be t hours.

60t + 40 = 50 (t + 3)
60t + 40 = 50t + 150
10t = 110
t = 11 (Speed of the faster car)

Therefore t + 3 = 14. (Speed of the slower car)

Need massive help:



I really have no idea. Yet it seems so simple.

let x be the time taken to jog one lap.

The time taken to walk one lap = 2x, since it's at half speed.

Total time is therefore 8x+2x=10x=30 minutes
x= 3 minutes = 0.05 Hours

5km/h x 0.005 hours = 0.025 km = 250 m.

Yawn.

Emma was too busy hovering around trying to make sparks with electricity in science today...what's the difference between the electrolysis of a copper substance using carbon electrodes, and the electrolysis of a copper substance using copper electrodes?

cookies to whoever can help me out before tomorrow!

11^(x+7) = 2^(7x)
11^x * 11^7 = (2^7)^x
11^x * 11^7 = 128^x
11^7 = 128^x / 11^x
11^7 = (128 / 11)^x
7 log (11) = x log (128 / 11)
x = 7 * log (11) / log (128 / 11)

x ~ 6.83958583

x isn't a very nice number -- either you wrote down the question wrong, or someone was being evil when they wrote it.

Hey there, I'm back on the forums! It's been some break.



I'm studying to be a biochemist, so I hope I can work this out.

Basically the difference would be (besides basic experimental differences) is that the end result in the first one is a copper substance covered in carbon, while in the second, the result is a copper substance covered in another layer of copper.

I don't have the time (nor the immediate knowledge) to work out any experimental differences.

I hope I get shot for asking a maths problem on here xD I never usually get help with maths.
Okay, for my upcoming GCSEs, I have been given past papers (higher tier) to work from. Around question 21 (2nd to last question), I've spotted similiar questions on all the papers, and should know how to deal with them before the 2007 exam :p
Here's the question:

Simplify fully: (4x^2-6x)/(4x^2-9)

Can someone please tell me the process in how to get the answer, because it's driving me insane There's other questions such as simplifying (x^2-3x)/(x^2-8x+15), which I think requires the same process :p

(4x² - 6x) / (4x² - 9)

Simplifying the Numerator:

Take 2x out of the brackets => 2x (2x - 3)

Simplifying the Denominator:

Factorise it to give you (2x + 3) (2x - 3)

You'll be left with:

(2x (2x - 3)) / ((2x + 3) (2x - 3))

Cancel out the (2x - 3) and you're left with 2x / 2x + 5.

Same thing with the other one

Of course *shoots* xD Thanks for the help ^^ I'll try and do the other then

(x² - 3x) / (x² - 8x + 15)

Factorise the numberator to get:

x(x - 3)

Factorise the denominator to get:

(x - 5)(x - 3).

Cancel out the (x - 3)s to get x/x-5 ?

Exactly

aaah, on the same paper there is this question which is the last one I need to do:

Here is a sketch of the graph y = 25 - (x - 8 )² /4

etc... the graph is shown...
P and Q are points on the graph
P is the point at which the graph meets the y-axis
Q is the point at which y has a maximum value

Find the co-ordinates of P and Q.
P is obviously (0,something) and y (8, something) But I don't know the somethings

and part B:
Show that:
25 - (x - 8 )² /4 = (2+x)(18-x)/4