# Question game

Hii everyone. This note is a type of quiz game where you can solve questions and post them too. The advantage of this note is you can practice both solving and posting good questions. Your solutions and questions will be appreciated too by community members. So, enjoy the beauty of solving the questions. Go ahead :

Rules And Regulations :

1. In this game the questions will be posted and by the community members itself.

2. The one who solves the latest question first will be able to post a new question. You should provide a clear solution to the question you have solved. Only the one who first solves the latest question correctly will have a chance to post the new question.

3. If any member has a different solution to a question which has already been solved he can also post the solution but he cannot post a new question.

4. There is also a leader board (below) in which the top 10 highest solvers will be shown. So try to solve as many questions as possible.

5. The one who posted the question should not be the first one to solve that question. However he can post his solution if he wants only after anybody solves his question.

Summary

Total Questions Solved : 6

Latest Question Posted : Problem 6

Leader Board (Top 10 highest Solvers) :

NAME - PROBLEMS SOLVED

$1$.Vilakshan Gupta - 2

$2$. Ram Mohith - 2

$3$. Vaibhav Priyadarshi and R Mathe and Michael Mendrin and Tom Clancy - 1

$4$. Matin Naseri - 1

If any improvements are required in this game kindly inform them to me.

Thank You and All the Best

Note by Ram Mohith
1 year, 4 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

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Note: you must add a full line of space before and after lists for them to show up correctly
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Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
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2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

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This is a nice discussion. Really enjoyed

- 1 year ago

Problem 1 (solved by Vaibhav Priyadarshi)

Find the value of $\log_2 10 - \log_8 125$

- 1 year, 4 months ago

$log_{2}10-log_{8}125$

=$log_{2}10-log_{2^3}125$

=$log_{2}10-(log_{2}125)/3$

=$(3log_{2}10-log_{2}125)/3$

=$(log_{2}10^3-log_{2}125)/3$

=$(log_{2}(1000/125))/3$

=$(log_{2}8)/3$

=3/3 =1.

- 1 year, 4 months ago

Now can I post a new question?

- 1 year, 4 months ago

Yes you can post a new question

- 1 year, 4 months ago

- 1 year, 4 months ago

Another solution for Problem 1 :

$\log_8 125 = \log_{2^3} 125 = \dfrac{1}{3}\log_2 125 = \log_2 \sqrt[3]{125} = \log_2 5$

$\log_2 10 - \log_8 125 = \log_2 10 - \log_2 5$

$\implies \log_2 \dfrac{10}{5} = \log_2 2 = 1$

- 1 year, 4 months ago

$\textbf{Problem 2 :}$

Find the value of $(5+2\sqrt{13})^{1/3}+(5-2\sqrt{13})^{1/3}$

- 1 year, 4 months ago

$\frac{1}{4}(1+3\sqrt{13}+i(\sqrt{39}-\sqrt{3}))$?

- 1 year, 4 months ago

Sir, if this is correct you can post a new question but whether it is correct or not Vibhav should tell it.

- 1 year, 4 months ago

I don't think the answer is "1"

- 1 year, 4 months ago

Just manipulate the expression using algebra and you will see that...wow... it turns out to be 1!

Some things look complicated but just do the math and you will get the answer.

- 1 year, 4 months ago

I will give you a hint , note that $a^3+b^3=10$. Rest is upto you

- 1 year, 4 months ago

Vilakshan Gupta why can't you post your solution if you have solved

- 1 year, 4 months ago

$(a^3+b^3)=(a+b)(a^2-ab+b^2)=10$, but I don't see where the "1" comes up. Both factors are complex conjugates.

- 1 year, 4 months ago

Actually what I meant was Vaibhav's solution only, because if you cube the expression, the radicals get cancelled off.

If you let the expression to be equal to $x$ and then cube it, then you will get the equation $10-9x=x^3$ and from here, by observation , we can conclude that the answer has to be $1$.

The other two roots have to be complex conjugates, but the answer in real numbers is $1$.

- 1 year, 4 months ago

Ok. Now I understood. So you have solved it Vilakshan Gupta

- 1 year, 4 months ago

Can you give explanation for it? Why isn't this a real number? Principal Cube root of real number should be real.

- 1 year, 4 months ago

First tell is it correct or not

- 1 year, 4 months ago

A lot of algebra went into this, but i can't get rid of the complex part.

- 1 year, 4 months ago

Let me try another way to do this

- 1 year, 4 months ago

I think not.

- 1 year, 4 months ago

- 1 year, 4 months ago

Yes it is correct!

- 1 year, 4 months ago

How is it "1"? This expression is a root of a 4th degree polynomial with no real roots.

- 1 year, 4 months ago

Let $(5+2\sqrt{13})^{1/3}$ + $(5-2\sqrt{13})^{1/3}$ = $x$

Cubing both sides, we get

$10+3(5^2-(2√13)^2)^{1/3}x$ = $x^3$

$10-9x=x^3$

It have 1 as a solution.

- 1 year, 4 months ago

Why have you posted the solution when you have posted the question

- 1 year, 4 months ago

Sir have asked me how it is 1, so I have given the reason.

- 1 year, 4 months ago

But the problem now is who solved this question and who will post the next one.

- 1 year, 4 months ago

Vilakshan Gupta

- 1 year, 4 months ago

Also I didn't understand how you got 9x

- 1 year, 4 months ago

By using $a^3+b^3+3ab(a+b)$

Here a+b=x.

- 1 year, 4 months ago

What about ab. When they both are multiplied they give a complex number

- 1 year, 4 months ago

$(5^2-(2\sqrt{13})^2))^{\frac{1}{3}}$ is complex. You slipped up on sign. Or at least about fractional powers of $-1$.

Let me try to see if I can save this....

- 1 year, 4 months ago

Is $(-27)^{1/3}$ not real?

- 1 year, 4 months ago

Unfortunately, no. This is where it gets complicated.

- 1 year, 4 months ago

- 1 year, 4 months ago

I agree that the principal root of $(-27)^{\frac13}$ is an imaginary number , but we are concerned about the real valued root , and that is , $-3$.

Check this

- 1 year, 4 months ago

Vaibhav shall I update the note as Vilakshan Gupta solved your question. What do you think?

- 1 year, 4 months ago

Consider an arbitrary polynomial $X^{3}+aX+b\in\mathbb{R}[X]$. Then, provided $\Delta:=(a/3)^{3} + (b/2)^{2} > 0$, letting

$C:=-b/2+\sqrt{\Delta},\quad D:=-b/2-\sqrt{\Delta}$

one has (taking real cubed roots) that $C^{1/3}+D^{1/3}$ is the unique real root of $X^{3}+aX+b$.

Now choosing $b:=-10$ and $a:=9$, one has $5+2\sqrt{13}=5+\sqrt{52}=C$ and $5-2\sqrt{13}=5-\sqrt{52}=D$ (one may obtain the values of $a,b$ by simply solving $-b/2=5$ and $\Delta=52$. Thus the expression in the problem, $\breve{x}:=(5+2\sqrt{13})^{1/3}+(5-2\sqrt{13})^{1/3}$, is equal to the unique real root of

$X^{3}+9X-10$

Now, note that $1$ is a root of this polynomial. Thus it is the unique real root. Hence $\boxed{\breve{x}=1}$.

- 1 year, 4 months ago

Good. Now you too solved this question but you cannot post the new question because Vilakshan Gupta already solved it. But this solution comes into your account.

- 1 year, 4 months ago

If we rewrote the expression to $(5+2\sqrt{13})^{\frac{1}{3}}-(-5+2\sqrt{13})^{\frac{1}{3}}$, then, yes, the value becomes $1$. But what is done is to assume that $(-a)^{\frac{1}{n}}=-(a)^{\frac{1}{n}}$, which is not true for all $n>1$. That's not a safe step to take when doing computations, because what that does is to "convert a complex quantity into a real".

This is a controversal subject, involving fractional powers of complex numbers, which are multi-valued, and this sort of thing crops up on Brilliant from time to time.

- 1 year, 4 months ago

@Ram Mohith But this solution comes into your account. I have no idea what this sentence means. Regardless. I solved it for the sake of solving it. I don’t pay attention to who does what first nor do I look at other people’s solutions until I’ve done things myself.

@Michael Mendrin I worked completely in $\mathbb{R}$. For odd $n\in\mathbb{N}$ the unique real $n$th root of $-a$ does coincide with $-(a^{1/n})$. I do appreciate that if we work in $\mathbb{C}$ it is all a matter of convention: there is no compelling reason to choose one argument over another (ie the choice of $k\in\mathbb{Z}$ in $\frac{\theta+2\pi k}{n}$).

- 1 year, 4 months ago

Well, right, what do we do if we have multiple values for this sort of thing? Just like just because we know that $(-2)(-2)=4$ doesn't mean it's safe to assume that $\sqrt{4}=-2$. Likewise fractional powers of negative numbers. Whenever I see something like that arise, it's never safe to "convert a complex into a real".

What's the value of $(-1)^{\frac{1}{3}}$?

I meant to explore this more in detail this morning, but this Question Game moves on.

- 1 year, 4 months ago

The value of $(-1)^{\frac{1}{3}}$ should be $(-1)^{\frac{1}{3}}$, which denotes a multiple-valued complex quantity. It's kind of analogous to some quantities in quantum physics, huh? "It can have a range of values, depending on how you look at it".

- 1 year, 4 months ago

It is not a multivalued object (except maybe in set theory). In algebra, it’s just ill-defined. And in complex analysis one acknowledges the existence of many inverses, and the consequences of this being any choice of the $n$th root function $f:z\in\mathbb{C}\mapsto \sqrt[n]{|z|}\exp(\imath\frac{\arg(z)+2\pi k}{n})$ is holomorphic except along a semi-infinite line.

- 1 year, 4 months ago

"In algebra, it' s just ill-defined" (!). Well, isn't that what I've been trying to argue?

Why is my value $\frac{1}{4}(1+3\sqrt{13}+i(\sqrt{39}-\sqrt{3}))$ any less valid than "1"?

- 1 year, 4 months ago

Being ill-defined here means there be no single-valued (inverse) function $f:\mathbb{C}\longrightarrow\mathbb{C}$. Being multi-valued means a function of the form $f:\mathbb{C}\longrightarrow\wp(\mathbb{C})$. Both of these are technically true. But in algebra, it is not sensible to work with the latter.

Your result is fine. The whole point of this discussion, was you disputing mine. I defended my result by stating that I’m working in the framework $\mathbb{R}$. You’re working in the framework $\mathbb{C}$. Which framework one chooses is another issue. As soon as these are fixed, the answers are what they are—within the framework $\mathbb{R}$ the only valid result is $1$. In your framework, which is perfectly legitimate your result is one of several perfectly valid answers.

- 1 year, 4 months ago

Oh, I don't know how you got the idea that I was disputing your solution. My apologies. Yes, earlier, before you jumped in with your solution, I did blurt out, "let's make sure if 1 is right!" Well, it was late at night, and so I didn't get the chance to review the matter until this morning, and saw what the problem was.

This is not the first time I've run to this sort of thing in Brilliant.org, where powers of complex numbers are involved. Several problems have already been posted to explore this topic.

An annoyance that I have is that for the sake of rigor, a function of the form $y=f(x)$ is, by convention, defined to be single valued. This is the reason why $y=\sqrt{x}$ by convention only yields positive $y$, i.e. only the top half of the parabola. But lots of implicit functions of the form $f(x,y)=0$ are multi-valued in both the $x$ and $y)$ directions. This is the reason why whenever complex quantities are involved, especially powers of them, it raises red flags for me and I move with caution, and ask myself, "okay, what's the convention with this, and in what context is this? " This is just from habit now.

- 1 year, 4 months ago

Sir, please try to solve the latest problem 3

- 1 year, 4 months ago

Not a problem. Generally any randomly chosen implicit functions will have many locally well-defined branches. There’s no mathematical preference for any. By contrast there is sensible preference however, when one demands restricting a solution to say a particular field. Nonetheless, as I said, one always needs to fix a framework and work in it. If things happen to have more solutions in a larger framework, then so be it. For example asking, whether $X^{2}+X+1$ be irreducible without specifying an algebraic structure, is simply begging for problems, as the answers is yes or no, depending on the context (if the field is $\mathbb{R}$, $\mathbb{Q}$, $\mathbb{C}$, $\mathbb{F}_{2}$, … or if one works in a ring, etc.).

- 1 year, 3 months ago

Your example with even values is irrelevant to what I was saying. I referred only to odd $n\in\mathbb{N}$. For odd powers there is only one real root.

For $n\in\mathbb{N}$ odd, the map $f:x\in\mathbb{R}\mapsto x^{n}\in\mathbb{R}$ is continuous and unbound above and below, thus the image (as a continuous image of a connected set) is all of $\mathbb{R}$, moreover the map is injective. Hence there is a unique Inverse in $\mathbb{R}$. Thus for all $a\in\mathbb{R}$ there is exactly one value $b\in\mathbb{R}$ such that $b^{n}=a$. This is the unique real $n$th root. Eg, the unique real $3$rd root of $-1$ is $-1$.

it's never safe to "convert a complex into a real"

I disagree with this. What is sensible is it is never safe to not fix a framework. And fix a framework I did. There is no compulsion to work in one particular field. Since eg $\mathbb{Q}\subseteq\mathbb{R}\subseteq\mathbb{C}\subseteq\mathrm{GL}(\mathbb{C},3)\subseteq\ldots$ are field extensions, one could by your logic also argue, that it’s never safe to work in $\mathbb{C}$ and that one should always work in some $\mathrm{GL}(\mathbb{C},3)$.

- 1 year, 4 months ago

Let me try to explain what I think the problem is. Suppose instead of something as complicated as the original expression, the problem asks, "what's the value of $(-1)^{\frac{1}{3}}$"? There is no one unique answer, unless the problem asks for the unique real root. If one argues that "1" is a valid value of the original complicated expression, I can equally argue that my complex value is equally valid.

What I am saying, in ordinary mathematical computations, one must be careful about assuming only the unique real roots of such fractional powers of negative quantities. Otherwise, I could "show" that $-1 = \frac{1}{2} +\frac{\sqrt{3}}{2}i$, for example (and therefore -1= 3!)

- 1 year, 4 months ago

It's creative. I want to join into your discussion.

- 1 year, 4 months ago

You can. In fact I am encouraging everyone to join this. Try to solve any question the first and you can post a new question.

- 1 year, 4 months ago

$\log_2 10 - \log_8 125$

$\log_8 125 ={\log_{2^3} 125}$

$\dfrac{1}{3}{\log_{2}125}$, $\sqrt[3]{125}=5$

Thus $\log_2 5 ={\log_8 125}$

$\log_2 10 -$$\log_2 5 =$$\log_2{\frac{10}{5}}=$$\log_2 2 = 1$

- 1 year, 4 months ago

Fine but you are the third solver to that question

- 1 year, 4 months ago

What I should do?

Also I have fixed my code,My last line had Errors.

- 1 year, 4 months ago

No problem. Leave your solution. It will come into your account. If you solved another question you will have 2 solutions and will be top in the leader board. Try to solve the latest question : Problem - 3

- 1 year, 4 months ago

$\text{Problem 3:}$

In the $\triangle ABC$ ,$O$ is any arbitrary point in the interior of the triangle. Lines $OA$ , $OB$ , $OC$ are drawn such that the angles $OAB$,$OBC$ and $OCA$ are each equal to $\omega$ , then prove that $\cot A+\cot B+\cot C=\cot \omega$ and $\csc^2A+\csc^2B+\csc^2C=\csc^2\omega$

- 1 year, 4 months ago

Ok. It is tough but no problem. But from next time onwards try to post problems of Easy and Medium difficulty.

- 1 year, 4 months ago

Let $a,b,c$ be the angles of the triangle and $x$ be the angle common to all 3 vertices. Then using Law of Sines, we have

$\dfrac{Sin(x)}{OB}=\dfrac{Sin(b-x)}{OA}$
$\dfrac{Sin(x)}{OC}=\dfrac{Sin(c-x)}{OB}$
$\dfrac{Sin(x)}{OA}=\dfrac{Sin(a-x)}{OC}$

Hence

$4{ \left( Sin(x) \right) }^{ 3 }=4Sin(a-x)Sin(b-x)Sin(c-x)$

And from there on, using product-to-sum and sum-to-product trig identities, we have

$3Sin(x)-Sin(3x)=Sin(2a+x)+Sin(2b+x)-Sin(2(a+b)-x)-Sin(3x)$

then after rearrangement and adding the term $-Sin(2a-x)$ on both sides

$-Sin(2a-x)+Sin(x)+Sin(2(a+b)-x)-Sin(2b+x)=-Sin(2a-x)-2Sin(x)+Sin(2a+x)$

then

$-4Cos(x)Sin(a)Sin(b)Sin(a+b)+4Cos(a)Sin(b)Sin(a+b)Sin(x)=$ $-4Cos(b)Sin(a)Sin(a+b)Sin(x)+4Cos(a+b)Sin(a)Sin(b)Sin(x)$

and finally

$Sin(a)Sin(b)Sin(c)Sin(x)(Cot(a)+Cot(b)+Cot(c)-Cot(x))=0$

- 1 year, 3 months ago

Good solution sir. If it is correct you are the one who can post the next question. But let's see what Vilaskhan Gupta says.

- 1 year, 3 months ago

Can you explain your first step, from where did the first equation come?

- 1 year, 3 months ago

I'll put that first step in, see my solution

- 1 year, 3 months ago

Vilakshan Gupta is Micheal Mendrin solution correct or not. Please tell I have to update the note.

- 1 year, 3 months ago

Ram, the full solution is going to be much longer than what I've posted, many more steps inbetween. What I've posted are the key steps, especiallly the middle part. I've not been able to figure out a shorter way to prove it. Maybe Gupta can just post a new question and start over.

- 1 year, 3 months ago

The solution seems fine, mine is also a similar solution. Then according to the rules of this game, you can post a new question Sir.

- 1 year, 3 months ago

Ok.Now the question is solved by Michael Mendrin sir and he can now post the latest questions.

- 1 year, 3 months ago

Two unit disks are randomly thrown on the plane. If they overlap, what's the expected distance between their centers?

- 1 year, 3 months ago

$\text{Problem 4:}$

Two unit disks are randomly thrown on the plane. If they overlap, what's the expected distance between their centers?

- 1 year, 3 months ago

1

- 1 year, 3 months ago

Ah, that's what's wrong with this problem, if I say your answer is wrong, then somebody else will probably figure out the correct answer.

- 1 year, 3 months ago

I solved it like this: If they overlap, maximum distance between their centres would be 2(if they just touches each other). Minimum distance between their centres would be 0, when they are coincident. So expected distance can be average of maximum and minimum which is 1.

- 1 year, 3 months ago

Okay, well, no, 1 is not the correct answer. In fact, if you do a computer simulation, randomly throwing unit discs on the plane, you won't get 1.

- 1 year, 3 months ago

The answer is $\boxed{0}$.

- 1 year, 3 months ago

The minimum and maximum possible distances are 0 and 2, so the expected value is somewhere inbetween.

- 1 year, 3 months ago

Okay, thus I think the final answer is $\boxed{0.5}$.

What about $\boxed{0.5}$?

- 1 year, 3 months ago

In mathematics, you don't get to guess. But, I can tell you it's some kind of a fraction.

It's not 0.5 and it's not 1/2

Pick a fraction between 0 and 2.

- 1 year, 3 months ago

I know two methods of doing it, but I know none!

One method is by using an excel sheet and other is by integration (both of which I don't know)

Just another guess....0.67

- 1 year, 3 months ago

Yes, the correct answer is 2/3

- 1 year, 3 months ago

Sir,you may ask another question since this was solved by none...

- 1 year, 3 months ago

Sir, I will mark this as solved by Vilakshan Gupta. I am giving chance for Mohammad Farhan to post the latest problem.

- 1 year, 1 month ago

I think this game lacks number of solvers

- 1 year, 2 months ago

Yes, probably.

- 1 year, 2 months ago

76 members have seen this note but only 6 people are participating.

- 1 year, 2 months ago

Make it 7

- 1 year, 2 months ago

Are you interested in participating ?

- 1 year, 1 month ago

Yes!

- 1 year, 1 month ago

Ok young dynamite now you can post the problem 5. Before that do you know the regulations of this game. If not read them in the above note.

- 1 year, 1 month ago

Thank you.

- 1 year, 1 month ago

do you have any question ready

- 1 year, 1 month ago

- 1 year, 1 month ago

You should give any link. You should write the question in a separate new comment. Also, try to post a problem which is not from community problems.

- 1 year, 1 month ago

Oh OK

- 1 year, 1 month ago

If many people start to participate then this game would be really creative.

- 1 year, 2 months ago

Yes,Maybe. but How to we should solve the issue?

- 1 year, 2 months ago

Do you have any idea ?

- 1 year, 1 month ago

- 1 year, 1 month ago

Probably. But the problem is all those who will see may not participate in this.

- 1 year, 1 month ago

Let's try that first

- 1 year, 1 month ago

Where is the 5th problem

- 1 year, 2 months ago

It is not posted till because nobody solved question 4.

- 1 year, 1 month ago

Problem 5: (Solved By Ram Mohith)

There were 5 friends namely Beatricia, Jing Kai, Emmanuel, Freddy and Benjamin.

Beatricia : 9 sweets

Jing Kai: 7 sweets

Benjamin : ? sweets

Emmanuel : 8 sweets

Freddy : 6 sweets

What is ?.

(LaTeX is hard)

- 1 year, 1 month ago

7 sweets or 5 sweets. I had both explanations. I you tell the correct answer I will post the relevant one.

- 1 year, 1 month ago

WRONG! YOU LOSE A LIFE! Sorry TWO LIVES! 1 MORE TRY

- 1 year, 1 month ago

Just Kidding

- 1 year, 1 month ago

The correct answer is $8$ sweets.

- 1 year, 1 month ago

Explain

- 1 year, 1 month ago

I have posted my solution. I forgot to tell one thing. Please write the question number (here Problem 5) at the starting of your question.

- 1 year, 1 month ago

OK

- 1 year, 1 month ago

Solution for Problem 5

The number of letters in each person's name is the count of sweets they have.

 Beatricia Jing Kai Emmanuel Freddy Benjamin 9 7 8 6 $\boxed{8}$

- 1 year, 1 month ago

That's Right! That's what the Gifted Education Programme in Singapore has

- 1 year, 1 month ago

You are now tied With Vilakshan Gupta

- 1 year, 1 month ago

But Vilakshan Gupta answer two questions which are not posted by himself. But I answered one question which I itself posted and now your question. So, first priority is given to him.

- 1 year, 1 month ago

OH

- 1 year, 1 month ago

Ram, You forgot to follow (Combinatorics) and (Electricity and Magnetism)

- 1 year, 1 month ago

Because those chapters are not started yet in our academics.

- 1 year, 1 month ago

Oh

- 1 year, 1 month ago

But do you know permutations?

- 1 year, 1 month ago

Yes some basics in our class 10.

- 1 year, 1 month ago

- 1 year, 1 month ago

$\text{Problem - 6}$

A ball falls on surface from $10~m$ height and rebounds to $2.5~m$. If the duration of contact with floor is $0.01$ seconds. Find the average acceleration during the contact ?

$(A) 2100~m/s^2$

$(B) 1400~m/s^2$

$(C) 700~m/s^2$

$(D) 400~m/s^2$

- 1 year, 1 month ago

Wow, you are even kinder. You gave options

- 1 year, 1 month ago

How many times can we pick the options?

- 1 year, 1 month ago

It not based on how many times you attempt. The first one who gives the correct solution with explanation is said to be as the first solver.

- 1 year, 1 month ago

(D) because 10/2.5 =4

according to acceleration = m/s^2

so 4/0.01=400

- 1 year, 1 month ago

No. wrong explanation.

- 1 year, 1 month ago

I think the correct answer is $2100{\frac{m}{s^2}}$ but I'm not sure about it.

Note: you have a typo in your first line. Find no fid

- 1 year, 1 month ago

Yes you are correct. But you should provide your explanation.

- 1 year, 1 month ago

Again I'm not sure about my solution.

$2.5×10=25;{\frac{25}{0.01}}=2500$

$10÷2.5= 4; {\frac{4}{0.01}}=400$

$2500 - 400 = 2100$

Hence the answer is $2100{\frac{m}{s^2}}$

- 1 year, 1 month ago

Okay but why you multiplied and divided 2.5 and 10

- 1 year, 1 month ago

Is my solution right?

- 1 year, 1 month ago

Yes it is correct. Tom Clancy you have solved this question so you can now post a new question.

- 1 year, 1 month ago

Okay, I will post my problem as soon as possible, but is it your own problem?

- 1 year, 1 month ago

Tom Clancy when are you going to post your problem.

- 1 year, 1 month ago

I have been waiting for a new question....

- 5 months, 4 weeks ago

@Ram Mohith, When will @Tom Clancy post a new question

- 12 months ago

Don't know. As he is a new user he came to brilliant only 2 or 3 days and then he stopped.

- 12 months ago

Maybe you post a new question

- 12 months ago

I am now in vacation and out of station and will not be available for 2 days more. So I cannot post a problem now.

- 12 months ago

- 12 months ago

@Ram Mohith, can you please post another question?

- 11 months, 2 weeks ago