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Huh? The x values differ in each equation.
And the most boring one would be Q
I don't get it?
Ah... this is a classic joke, I think I learnt it early in junior school.
Aww, I was only 6 letters out.
Oh noes, somebody contradicted themselves ![]()
Nope, nobody said anything.
If somebody is a nobody then they are both a somebody and a nobody.
Would that be somewhere to the east of the City of Rats?
But they will never exactly overlap one another, won't they?
It's still working fine for me
Welcome Fried! Hope you enjoy our discussions.
Like ur avi
i?
I think words starting or ending with 'I' sound cool. e.g. Icicle, Isotope, Gemini. Plus, it's the square root of -1, which is pretty kewl 2.
If the x value's tenth's column is greater than or equal to 5, then is x rounded up, or is it always rounded down?
But... there is nothing missing. You can't say that the smallest possible number is missing because there is no such thing. The concept of infinity eliminates it. Infinity is so infinitely enormous that it, by brute force, vanquishes the smallest possible number.
Oh...in my book it says
int(1000 ÷ 7) = 142
This is, in a way related,
Does 'int' in int(x) stand for integer? If it does then is x rounded up or down?
Thanks anyway pi man, I guess I got too used to doing it in one step, as I have learned to do with surds and trig to 'keep things accurate' ![]()
It would be nice if we had a weekly or even daily high-score board to encourage more use of this mathematical tool. Otherwise, once the 'eternal' highscores have been set, usage may stagnate.
Oh I see, 99 ÷ 4 = 24.75. Rounded down = 24. If I had seperately worked that out instead of using the combined sum that I used, I would have gotten 101 for my answer.
Thanks everyone.
Oh i see, so if I'm looking for integers between x and y, where x > 0, then I simply have to add one.
But what if I had a question, to say, find between 14 and 18 (inclusive) all integers divisible by 4:
(18 ÷ 4) - (13 ÷ 4) = 1.25. If I were to round this up as has been suggested it would be incorrect.
Oh noes it doesn't make any sense! ![]()
Yep, I usually find the first few questions are the slowest, and then the rest come much faster.
Question
How many numbers between 100 and 500 (inclusive) are divisible by either 4 or 5?
The correct answer is 161 numbers. I am out by just one, with an answer of 160. Can someone please correct my working so it gives the correct answer. Thanks.
Solution:
Let n(A) = numbers divisible by 4
Since (500÷4)-(99÷4) = 100.25, there are 100 numbers between 100 and 500 divisible by 4.
i.e. n(A) = 100.
Let n(B) = numbers divisible by 5
Since (500÷5)-(99÷4) = 80.2, there are 80 numbers between 100 and 500 divisible by 5.
i.e. n(B) = 80.
To find n(A and B), notice that if a number is divisible by both 4 and 5, it must also be divisible by 20.
Since (500÷20)-(99÷4) = 20.05, there are 20 numbers between 100 and 500 divisible by 20.
Using n(A or B) = n(A) + n(B) - n(A and B) gives
n(A or B) = 100 + 80 - 20
∴ n(A or B) = 160
Hence, there are 160 numbers between 100 and 500 divisible by either 4 or 5.
Well, seeing as complementary angles add up to 90 degrees and you have three times one of the angles, you can draw these algebraic conclusions:
1.
2.
You can now use simultaneous equations to find angle A.
EDIT: Guess pi man beat me to it ![]()