How do you solve this math question?
You need to use the concept of relative velocity to solve this problem. But before that I will take the liberty to add few details to the question. Let's say the difference between point A and point B is D. And for this journey to be feasible the bird should fly with a speed v > 10kmph .Scenario A. Point A is behind point BBird has started it's journey to point B at time t = 0. It reaches the point B at time t1. Which is given by the formulaTime = distance / speedNow when bird is flying from point A to point B with a speed of v. The B is also moving ahead of the bird with a speed of 10 kmph. So when bird covers v km in an hour. Train has moved forward by 10 km. So the distance to point B has been reduced to V - 10 km. In other words velocity of bird relative to train while the bird flies towards point B is vr1 = v - 10 So t1 = D / ( v - 10 )When bird is flying back to point A. The train is also moving towards the bird with a speed of 10kmph. So vr2, that is the velocity of bird relative to point A is v + 10Therefore t2 = D / ( v + 10 )So total time t = t1 + t2 = 2Dv/ ( v^2 - 100 )Scenario 2: point A is ahead of point BThe scenario is are reversed but you get the same answer t1 = D / ( v + 10 )t2 = D/(v-10)t = 2Dv/(v^2 - 100 )Substitute the values from the book for v and D and you get the answer. Please let me know if you need any more clarification on my answer Note: if the question is about the person walking inside the train instead of bird flying over it. The answer will be entirely different.
Can you solve this math question.............16 ÷ 2 [8 – 3 (4 – 2)] + 1?
16 ÷ 2[8 - 3(4 - 2)] + 1 = 16 ÷ 2[8 - 3(2)] + 1 = 16 ÷ 2(8 - 6) + 1 = 16 ÷ 2(2) + 1 = 16 ÷ 4 + 1 = 4 + 1 = 5
Can you solve this math question?
I shouldn't have watched that horror movie last night on TV. It gave me the strangest nightmare. I dreamed I was trapped on a planet of giant insects with Mega-Flies as big as elephants. I thought it was weird that the female mega-flies had 8 legs & the male mega-flies had only 3 legs. In my dream I was caught in the web of a Mega-Spider who would not let me go unless I could solve this problem: How many Mega-Flies would he need to catch if he eats 140 Mega-fly legs each day? What are all the possibilities?
Can you solve this math question?
I would choose the million dollars up front because I realize that there are not enough pennies in the world to pay me. There probably is not enough wealth in the world to pay me either by the penny method. Mathematically, the second question is answered by realizing that the summation yields 2^(365)-1 cents. As an interesting sidenote, the person that would "retire" after one month is very unrealistic since the day he retires he is turning down over $5 million for that ONE DAY. Not to mention he probably has spent quite a bit and may even be in debt by this thirty days. No one turns down a $5 million pay day, especially when the following day would be a $10 million pay DAY and so on. The amount of money you would have at the end of the year is mathematically calculated out to : US Dollars = ( (2^365) -1 ) / 100 approximately 7.515336264 x 10 ^107 dollars. Since, you could not possibly spend all of this money and the investment potential is almost unlimited who knows how much money someone could actually have. With this kind of money you could give each person in the world a trillion dollars a nanosecond for a trillion years and still not run out of money even without any interest. I guess I would make Bill Gates look quite stingy.
Can you solve this maths question?
#2. [math] ^nC_r + ^nC_{r-1} \\= \frac{n!}{r!(n-r)!}+\frac{n!}{(r-1)!(n-r+1)!}[/math]As we know,[math]r! = r \times (r-1)!\\ and, \space (n-r+1)! = (n-r+1) \times (n-r)![/math]So, LCM of r! and (r-1)! is r!And (n-r)! and (n-r+1)! is (n-r)!Using this, the LCM of [r!(n-r)!] and [(r-1)!(n-r+1)!] is [r!(n-r+1)!]Now,[math]= \frac{n!(n-r+1)+n!r}{r!(n-r+1)!}\\= \frac{n!(n-r+1+r)}{r!(n-r+1)!}\\= \frac{n!(n+1)}{r!(n+1-r)!} \space [\space n-r+1=n+1-r]\\= \frac{(n+1)!}{r![(n+1)-r]!}\\= ^{n+1}C_r[/math]Proved...I am trying to solve the others (honestly, I'll try when I get time).
Can you help me solve this math question?
Time x speed = Distance 3 x 100 000 000 m/s x time = 108 000 000 km 3 x 100 000 000 m/s x time = 108 000 000 000 m time = 108 000 000 000 / 3 x 100 000 000 m/s time= 360sec time= 6 mins
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Using the numbers 1-9, fill in the blanks to make AS MANY possible solutions that are true. Like this: 333 +333 ----- Except the 3's are the blanks. (this is an addition problem) There are 140 solutions in all, but I can't find all of them. There are 280 if you flip around some of the questions.
How would you solve this GCSE math question?
Umm, maybe if you specified what your mathematical problem is?
Math Question: Can you solve this Algebra question?
I am taking a survey of how smart the average American is for a school project. Can you completely factor this problem? 3*x*y^3 + 2*x^3*y - (7*x^2*y^2) Please be sure to include your answer along with how long it took you to solve it. Thanks!