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i saw this picture over social network
put into equation that will be:
Given:
A=rabit
B=Dog
C=Cat
A+C=10
A+B=20
C+B=24
solve A+B+C = ?
some programming logic? what is your answer?
I do not see any programming logic in this, its just Math.
B + C = 24
-A-C = -20
Adding these two,
B-A = 4
Now taking the 1st equation,
A+B = 10
-A+B = 4
Adding both again
2B= 14
B = 7
Then, A = 3 ( from A+B = 10) and C = 17kgs (from B + C = 24)
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Well, I have a little problem I want to solve.
We have two values. The first can go from 100 to 75. The second can go from 50 to 40.
How can I make it so when the first value is 100, the second is 50,
and when the first value is 75, the second is 40?
(For example 87.5 should be 45 on the second value)
I am trying to find that number.. Please give me the "formula" to do so.
Linear Interpolation
0.4 · Input + 10 = output
eg:
0.4 · 87.5 + 10 = 35 -10 = 45
number2 = ( ( (number1 - 75) / 25 ) * 10 ) + 40
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I always have trouble with factorials. Can someone walk me through simplifying this expression?
(x+1)! - 1 + (x+1)(x+1)!
I'm trying to get it to equal to (x+2)! - 1.
It can be solved as :-
(x+1)! - 1 + (x+1)(x+1)!
= (x+1)! + (x+1)(x+1)! - 1
= (x+1)!.{1+(x+1)} - 1
= (x+1)!.{x+2} - 1
= (x+2)! - 1. // since n!.(n+1) = (n+1)!
Hence proved.
Note that (x+1)! = (x+1)*x!
(x+1)! - 1 + (x+1)(x+1)!
= (x+1)!((x+1)+1) - 1
= (x+1)!(x+2) - 1
= (x+2)! - 1
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I'm trying to calculate AB+C (A, B and C being boolean values) but I only have the values ABC and C. I can negate, OR, and AND to my heart's content, but I am only getting ABC and C out of the system.
Specifically there is a service that has two boolean state variables. One is a combination of three internal properties and is only true when all three are true, but one of the three state properties is also available. I'm just trying to figure out if I can calculate the value I want without asking for them to add A and B as separate values.
AB is A and B
A + B is A or B
I don't think you can compute AB+C in any case knowing only the value of C and ABC:
If C is true, AB+C is true
If C is false, ABC is false and you can't find out anything about AB.
Old Answer - which is wrong as stated in the comments:
I'll give it a try. You can construct not(not(ABC)+C)+C:
not(not(ABC)+C)+C = not(not(A)+not(B)+not(C)+C))+C
= not(not(A)+not(B))+C
= AB+C
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limit x-->3 2x^2 + 7x-15/x-3
What i Simplified
Step 1 : 2x^2 + 10x -3x -15 / x-3
Step 2 : 2x( x + 5)-3( x + 5)/x-3
Step 3 : (2x - 3)(x + 5)/x-3
but unable to move further.. im thinking either the question ix wrong or there ix some trick which im unable to understand
thanx in advance
as x -> 3, the numerator goes to 3 * 8 = 24 and the denominator goes to 0, so the limit goes to +infinity if you approach 3 from the right, and -infinity if you approach 3 from the left
since you didn't specify which direction, the limit does not exist.
try graphing it: https://www.desmos.com/calculator
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Given N = A%B, how to find the value of A%C , where B > C.
You are given value of N and C, but not of A.
Is there any way to find this?
Nope. Consider the following:
A = 19
B = 10
C = 7
==> Given 9, you should get 5.
A = 29
B = 10
C = 7
==> Given 9, you should get 1.
So given the same input, there may be multiple answers.
The modulo operation is one-way: given a mod b = n, all I can say is that a comes from the set of all other integers which, modulo b, equal n.
Let's demonstrate that this is impossible in general, taking B=3, C=2.
n = a mod 3 = 1
=> a is in the set of integers {3x + 1}
so consider, x=1
4 mod 3 = 1, so that works
4 mod 2 = 0
now consider x=2
7 mod 3 = 1, so we can't distinguish 4 from 7 knowing only n and b
7 mod 2 = 1
That is, given b=3 and n=1, you'd have to get two different answers without knowing a.
However, you may consider it's a special case that b and c here are coprime, and in fact are both prime. You can certainly solve this easily for some cases, such as b=4 and c=2.
BTW, further discussion on this is probably better suited to mathoverflow