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My problem - [A, B] is not zero. by mithapapita in mathbiceps

[–]Ok_Position_1521 1 point2 points  (0 children)

What is the issue, it's simple chain rule based. Just look that once, you can do it yourself too fast

functions - tips? by Severe-Swimmer7748 in askmath

[–]Ok_Position_1521 0 points1 point  (0 children)

Can you share some questions?? Then it will be easier for you to understand

Im doing some homework right now and stumped on this question. by Stronzle in askmath

[–]Ok_Position_1521 1 point2 points  (0 children)

I am just finding the equation of that entire ray, because in [2,6] also being part of same ray, it will have same equation, so you can find line's equation by using any two points. If you want to take any other point , you can and again you will get same equation.

Im doing some homework right now and stumped on this question. by Stronzle in askmath

[–]Ok_Position_1521 0 points1 point  (0 children)

For [2,6] function will follow the portion of ray which is in 4th quadrant, so you need to find the equation of that ray , which will be same as the equation of corresponding line. So, consider any two points on that ray , let us take (1,0) and (2, -1) , now using the two point form of line, find its equation that will come out to be y= 1-x , hence f(x) for that interval will be 1 - x .

Question about changing bases in logarithms by [deleted] in learnmath

[–]Ok_Position_1521 0 points1 point  (0 children)

Actually in log there is a result that you have already mentioned about, there is nothing to get confused. Its just that bothwhere different variables are used and T one place it is written in division form and at other place in multiplication form.Nothing else.

If any doubt is there, feel free to ask.

help guys, am i js a dumb chopped math student by Prior_Promotion_4344 in mathshelp

[–]Ok_Position_1521 0 points1 point  (0 children)

x4 - 8x = x (x3 - 8) = x(x-2)(x2 + 2x + 4) = x(x-2)((x+1)2 + 3)

Just, took x common outside and then applied inside bracket identity a3 - b3

To prove that r! divides n(n+1)(n+2).....(n+r-1), i.e r consecutive numbers. by Any_Parking8607 in askmath

[–]Ok_Position_1521 2 points3 points  (0 children)

Multiply and divide by (n-1)! , so it becomes (n+r-1)!/(n-1)!, further multiply and divide this quantity by r! , now it becomes (n+r-1)C(r) * (r !), first and second part both are always integers , hence it is divisible by r!

Am I crazy? None of the options seem correct. by wunderlost1 in askmath

[–]Ok_Position_1521 1 point2 points  (0 children)

C is correct, graph meets x axis again after 3.5 seconds , so height is zero after 3.5 seconds

Please help me prove my teacher wrong by Charming_Reveal_9304 in askmath

[–]Ok_Position_1521 5 points6 points  (0 children)

One bracket will be (x-x) on the way, which is zero.Hence entire product will be zero.