Math Is Fun Forum

Ok I see now, it should be read as:
"Fifty minutes ago, if it was four time as many minutes past three o'clock" = 50 minutes ago, it was four times as many minutes (ie *4 times as much as that 50 minutes*) past 3 o'clock.

I agree now, this is probably how it should be interpreted. It's not as convoluted. But I still think it's a badly worded problem.

I may quite possibly be wrong, but I interpret the problem like this:

let the current time be T minutes. T is measured in minutes.

let the number of minutes be 0 at noon.

Therefore, T is the number of minutes that have passed since noon.

Writing 3PM in terms of minutes past noon:

3PM = 3 * 60 minutes = 180 minutes

Translating the given information into an equation:

T - 50 = 4(180 + T)

Solving for T:

T - 50 = 720 + 4T

3T = -770

T = -256.667

Writing 6PM in terms of minutes since noon:

6PM = 6 * 60 minutes = 360 minutes

let X be the amount of minutes needed to get to 6PM from the current time.

We know now that:

-256.667 + X = 360

X = 385.667

Therefore it is about 386 minutes to six o'clock.

In your case, you just have to re-arrange the equation so that it is in that form:

or equivalently:

I assume a is a constant. First factor it out:

Now we need to factor the numerator.

Call it f(r) = r^3 - r^2 + r - 1

Try values of r such that f(r) = 0. Notice that f(1) = 0.

Thus, by the factor theorem, (r-1) is a factor of f(r).

To find the other factor, use polynomial long division.

I get the other factor as (r^2 + 1)

Thus, f(r) = (r - 1)(r^2 + 1)

Now we can write:

I got this far but I get a different solution proceeding as follows:

We need to find the critical points of the expression, ie where it is zero or undefined:

it is undefined when x = 1 or x = -2.

Now looking at the signs of the expression on the intervals these conditions imply:

positive on

negative on

positive on

negative on

Therefore, the solution set is the union of the intervals

in other words, the solution set S = {x: x < - 2} UNION {x: 1 < x < 7}

You can always ask people on this site for help, some of us find it fun.

And yes I've had some lousy teachers myself in the past.

Can you give some examples of specific questions which you don't like or don't get? We'd be glad to explain anything !

This question is based on applying De Morgen ad nauseum, like 15 times:

For the 2nd, are you sure the answer is supposed to be:

~D + A~B~C + ~A~BC

I got ~D + A~B~C + ~A~BCD

ie, with a D at the end...

It's a pain to type though... I just applied De Morgen to ~(C+D), and then factored out ~D on every term possible, then what was left over simplified to 1, which left ~D.

I'll use "~A" to mean A negated.

We know that A + 1 = 1.

The OR operation needs *just one* of its operands to be 1 in order to evaluate to 1... so no matter what expression is "ORed" with 1 will give 1.

Therefore, ~A + 1 = 1, and AB + 1 = 1.

As for the second part:

Boolean And is commutative, so we can write:

A(~C) * (~A)(C) = A(~A) * C(~C) 

And in general, for any boolean X:  X * ~X = 0.

(ie, something can't be true AND not true at the same time!)

Therefore, A(~A) * C(~C)  = 0 * 0 = 0, as you said.

Notice that:

and also:

Thus:

Actually in my first year calculus course they also gave us limit questions which required L'Hopital's rule but which was not taught yet; I suppose there is some weird other way of doing it but this is the best way!

So here is L'Hopital's Rule for the case of rational functions:

Consider a limit

If f(a)/g(a) gives an *indeterminate form*, such as:

Then you are allowed to differentiate the top and bottom functions:

If this new limit also gives an indeterminate form, you can differentiate the top and bottom again until the quotient gives a 'normal' value.

(The actual definition is more technical and precise, but I think this is sufficient for school assignments )