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Note that

The two expressions have p^2 in it and 2ap if you put a = 7

So add a^2 = 49 and you have a perfect square.

As a general rule find half the middle coefficient (14/2 = 7) and square it (49).

Bob

We're in luck here because (x^3)^2 = x^6

So substitute Y = x^3 and you'll get a quadratic in Y.

Bob

hi CG

Welcome to the forum. At last I'm justified in keeping all my old pictures on my laptop.

This is what I have for that date.

Please confirm that's what you're looking for.

Bob

Nearly; but you have a 3 too many. In your box 3y times 3y doesn't make 3y^2.

Bob

to both of you:

I have edited the above post so that it displays properly. The correct command is square brackets math

When a member makes a post, it is visible to anyone who logs in to the site. And anyone can respond to any post. If someone disobeys our rules then I will take action. This may involve any of the following: Issuing a warning; Editing the offending part of a post; Deleting the post; Banning for a limited period; Removing the person completely from the membership. This last means that all that person's posts get deleted too.

At the moment I can see no reason to do any of these things. Let's keep it that way.

Best wishes,

Bob

You need to show that two of these when squared and added make the same result as the third when squared.

But which to choose? I can see that c is bigger than a. So I suggest working out c^2 and write that down. Then get an expression for a^2 + b^2. If this comes to the same as c^2 then you are done.

Bob

We seem to be getting abuse spilling over from another forum. I have removed the abuser. It will take me a while to fully clean up the unwanted posts. You can help by deleting any posts you have made that are now responses to nothing.

Thanks, Bob

Yes.

The number of noughts after the 1, and the power of ten amount to the same thing.

And if the power of ten is negative a similar rule applies.

If you would like a bit of extra practice here's some proofs you could try.

Show that the sum (and product and division) of two rationals is also rational.

Show similarly that two irrationals, when combined using + - x or ÷ is also irrational.

We say the rationals are closed under these four operations and the irrationals are similarly closed.

Are the rationals closed when we apply the square root function?

(Harder) Show that any rational is either a terminating decimal (eg 1/4 = 0.25) or a recurring decimal (eg. 1/3 = 0.33333....)

Hence deduce that any irrational must have an infinite, non recurring decimal form.

Bob

Most numbers (ie unless zero) have two square roots. The principal root is the positive one.

You posted a whole string of posts on the same subject. It has been hard for me to check if I've replied to them all. It would be easier if you either (i) waited for the first reply and then asked the next question as a further reply to that; or (ii) put all you questions in one post, numbering them so I can be sure I've got them all covered.

Bob

zero has no absolute value as it is not positive nor negative.

As he has already defined the principal root to be postive there is no need for the absolute to be stated again.

If you followed the other post on this you'll see we are just reversing those examples.

eg. 2.1 x 10^(4) = 2.1 x 10000 = 21000. The point has moved four places because that's what multiplying by 10000 does.

Because x could be negative and he is trying to define a unique positive root.

Sorry about the image problem. The old server did allow images to be upoaded but that was lost when MathsIsFun switched us to a new server. I use Imgur and uload images there; then link using bcc. There's a post about this here http://www.mathisfunforum.com/viewtopic.php?id=28084

Bob

Let a, b, c and d be integers.

ad - bc is an integer. bc is an integer. => the result of subtracting one rational from another is another rational.

Bob

When you change a number into scientific notation the actual value of the number must stay the same. The rule is to insert a decimal point between digits one and two and then correct the value by multiplying by a power of ten.

9582 = 9.582 x 1000 so the correct power is 3 (1000 = 10^3)

0.285 = 2.85 x 1/10 so the correct power is -1 1/10 = 10^(-1)

Back in the days when logs were used to do calculations it was an essential part of the process so that the table book only needed to have logs of numbers between 1 and 10.

Now with calculators it allows entry and display of numbers which would otherwise exceed the available display size.

Bob

Division is defined as the inverse operation to multiplication, so any division can always be re-written as a multiplication.

a/b = c may be re-written as a times c = b.

If we try to determine a numeric value for 0/0 re-write it as a multiplication:

If 0/0 = x then we want to find x so that 0 times x = 0. There is no single solution to this so x cannot be determined.

If a/0 = b then a times b = 0. The only solution for this is a = b = 0.

For all other values of 'a' no 'b' exists at all (different from the above as here we have no solutions whereas above we have multiple solutions.)

Bob

Prove this by assuming the converse.

Suppose a rational + an irrational is rational.

Re-arrange so that rational - rational = irrational. It is easy to prove that r - r is also rational.

Bob

No. It follows from what I have said in the other post.

Bob

hi harpazo1965

In a word, no.

The set of rational is all those numbers that can be written as the ratio of two integers:

eg 0.5 = 2/4

The set of irrationals is all those (real) numbers that cannot be so written.

eg √2.

There is a fairly well known proof that root 2 cannot be written as a ratio of integers.

Bob

In the first example the 75 angle is not in the 'same' position in both triangles. You have to expect this. Two triangles can be similar; if you turn one round a bit, they're still similar.

To work out which sides to use to make the ratio look carefully at the sides making the angle of 75. In one triangle it is 15 and 21; in the other it is 14 and 10. Now it must be the smaller sides that go to make a ratio, and the longer sides to make a ratio. Two triangles would never be similar if long and short were put together.

So make the ratios out of 15 and 10 for one; and 21 and 14 for the other.

15/10 and 21/14

In the second example we know the triangles are similar because they share the acute angle on the left and they both have a 90 angle. So similar AA.

Now to identify which sides in the red triangle have been scaled up to make the blue triangle. ? and 130 are the hypotenuse for each triangle. The sides adjacent to the acute angle are 80 and 127.

So think of the question like this. The 127 has been scaled down to 80. That's equivalent to multiplying 127 by 80/127.

So the unknown side, ? , must be scaled down by the same fraction:

? = 130 x 80/127

Hope that helps,

Bob

hi tokyokookie

Welcome to the forum.

With a question like this, you have to be careful about positive and negative.

Let's say we choose to make 'up'the positive direction. Then we have

s = - 117.6 because the ground is below the launch point.

u = + 24.5 as we are throwing upwards.

g = - 9.8 because gravity make things go down.

So using s = ut + 0.5at^2

- 117.6 = 24.5t - 4.9t^2

rearranging

4.9t^2 - 24.5t - 117.6 = 0

That's ok as you can multiply mine by -1 to get yours.

x by 10 and divide by 7

7t^2 - 35t - 168 =0

divide by 7 again

t^2 - 5t - 24 = 0

(t-8)(t+3) = 0 so t = 8 or -3

In the context of the question -3 isn't an answer so choose t = 8.

In fact -3 does have some significance. The ball obeys that equation even if we work backwards in time. If you start the ball on the ground at -3 seconds and let it zoom up, it will reach the clifftop at t = 0 and then carry on up for a while before coming back down at t = 8.

For part 20 just set s = 49. I haven't tried it yet but I'm expecting two answers; one when it's going up and another when it passes the same point on the way down.

Bob

hi Math Enthusiast

Just want to check I'm understanding the problem.

A triangle PQR with sides p, q, r (convention is p is opposite P etc)

Angle at P is 3Q => angle at R is 180 - 4Q

p, q, and r are all integers.

HCF(p, q, r) = 1

Find smallest side which is a perfect cube number {1, 8, 27, 64,.....}

Is that it?

Bob

hi PaulColbert

Welcome to the forum.

Bob