## How to convert a 1million+ decimal value to binary fast? Any technique? (without using calculator)?

We have a test tomorrow and we aren’t allowed to use calculators..please help..any fast technique?

If you mean a million digits then it would be hard to do on a test.

If you mean a number with a value over one million — seven digits then it is not too hard.

If the number is odd write down a 1 else write down a 0
Divide the number by 2 Drop any fractional part (integer divide)
if the new number is odd write down a 1 to the left of the previous 1 or 0
if it is even write down a 0 to the left of the previous number.

Repeat ‚til you get a zero after dividing by two

example
121
odd write 1
divide by 2
60
even we get 01
divide by 2
30
even we get 001
divide by 2
15
odd we get 1001
divide by 2
7
odd we get 11001
divide by 2
3
odd we get 111001
divide by 2
1
odd we get 1111001
divide by 2
0
stop

check: add 110 (6)
we get 1111111 (127) which is 6 + 121

## How do I determine the rational zeros of this function?

f(x) = -3x^4 + 4x^3 + 52x^2 – 64x – 64
My book only shows me one example of this and I can’t see how they did it because it is an even numbered problem and my book only shows solutions to odd problems. I’m just not sure how to get started on this type of problem. Usually if I see an example of how it is done, I can usually do them on my own from there. Thank you in advance for your help.

Rational zeros come by guessing and testing the possible rational zeros (obtained from the Rational Zeros Theorem):

Take the constant term and the highest degree term:

-3x⁴ and -64 in this case:

Forget about negatives, factor both of them, then take every possible combination of the constant factors over the highest degree terms factors, then try each of those by plugging them into the function, anything that gives 0 is a 0 of the function:

64 = 1 * 64, 2 * 32, 4 * 16, 8 * 8
3 = 1 * 3

So now REMEMBER it’s constant / highest degree term:

Try:
±1/1, ±64/1, ±2/1, ±32/1, ±4/1, ±16/1, ±8/1
and
±1/3, ±64/3, ±2/3, ±32/3, ±4/3, ±16/3, ±8/3

So now get your graphing calculator out and write the function in it:

y = -3x^4 + 4x^3 + 52x^2 – 64x – 64

Click graph, then choose trace, then enter in the x values one by one to obtain the following:

f(1) = -75.000000
f(-1) = 45.000000
f(64) = -49074240.000000
f(-64) = -51163200.000000
f(2) = 0.000000 <– x = 2 is a zero
f(-2) = 192.000000
f(32) = -2963520.000000
f(-32) = -3221568.000000
f(4) = 0.000000 <– x = 4 is a zero
f(-4) = 0.000000 <– x = -4 is a zero
f(16) = -168000.000000
f(-16) = -198720.000000
f(8) = -10560.000000
f(-8) = -10560.000000
f(1/3) = -79.444444
f(-1/3) = -37.074074
f(64/3) = -560305.777778
f(-64/3) = -635247.407407
f(2/3) = -82.962963
f(-2/3) = -0.000000 <– x = -2/3 is a zero
f(32/3) = -28811.851852
f(-32/3) = -37155.555556
f(4/3) = -56.888889
f(-4/3) = 94.814815
f(16/3) = -746.666667
f(-16/3) = -1277.629630
f(8/3) = 59.259259
f(-8/3) = 248.888889

So you see that the 4 zeros are:

x = 2, 4, -4, 2/3

Edit:

One thing you should realize about the Rational Zeros Theorem is that it ONLY guarantees to find all RATIONAL zeros…NOT all zeros of the function. What this means is that if we had "found" more than 4 zeros we know we made a mistake (since the maximum number of zeros for a degree 4 polynomial is 4…hence the quotes, because we wouldn't have actually found 4 zeros).

The point is, is that you might find LESS than 4 rational zeros, in fact you may find NO rational zeros!!!

The reasons this can happen are three-fold:

1. There could non-rational zeros
2. There could be complex zeros
3. There could be multiplicities (i.e. X² has a multiplicity @ x = 0 because x² = (x – 0)²

And of course you can get a combination of all three of those cases that leads to less than 4 (or whatever degree you have) of rational zeros.

## how do you calculate fractional betting odds?

I’ve been betting for years but don’t really know how to calculate fractional odds accurately. I can work out single bets ok, but, for example; how do you calculate a treble bet consisting of 6/4, 19/20 & 4/7. Show me.

Why wasn’t I taught this in Maths at school? Too busy learning those essential equilateral equations I suppose..

Really easy add the two numbers together and divide by the bottom number
so on a Calculator 10÷4×39÷20×11÷7=7.66x whatever stake you want gives the return for the treble

1 m+
10÷4 m+
39÷20 m+
11÷7 m+
14÷4×59÷20×18÷7 – mr x whatever stake you want gives you the returns for the 3 doubles and the treble

## How do you find the turning points for quartics?

a) y=x^4 -125
b) (x^2 – x -20)(x^2 -2x – 24)
c) y= x^4 – 4x^3
d) x^4 – 25x^2

is there a way to find the turning points without using the calculator please explain thanks.
a) y=x^4 -125x for a) it’s got an x on the back of 125 I forgot to put the x there

A. Y = x^4 – 125x = x(x^3 – 125) = x (x – 5) (x^2 + 5x + 25)

Real roots at 0 and 5. Imaginary roots for the quadratic, but there’s a turning point at -b/2a = -5/2. Also check halfway between the real roots, at x = 5/2. Looks to me that turning points are at +/- 5/2. Graph it using the factored version, and you’ll see …

B. (x^2 – x -20)(x^2 -2x – 24) = (x – 5) (x + 4) (x – 6) (x + 4)

A double root at x = -4 … Probably a turning point. Other roots at 5 and 6. Turning points likely at 1/2 and 5 1/2. Graph it using the factored version.

C. Y = x^4 – 4x^3 = x^3 (x – 4)

Triple root at origin indicates a saddle point (because the exponent is odd). Root at x=4, so there’s a turning point around x=2.

D. X^4 – 25x^2 = x^2 (x^2 – 25) = x^2 (x + 5) (x – 5)

Turning point at the origin (even exponent for double root). Look for two more turning points near x = +/- 5/2 (halfway between real roots).

## How to determine if the slope is negative or positive from two points?

The two points are (10,9) and (1,4)
4 – 9
1 – 10

=
-5/-9
Subtract

=
5/9
Simplify

The slope is
5/9

Thats what the online calculator for slope gave me. They change it from negative from positive. Why did they do that and how am i suppose to know when the slope is negative or positive for the two points? (finding the slope by using the formula)

Any time you multiply or divide two negative integers (such as -5, and -9), the two negatives cancel each other out, giving you a positive answer, in this case 5/9.

Just follow the order of operations, do the math, and you’ll know whether it’s positive or negative.

If, for example you were dealing with (10, -9) and (1, 4)

10-1 = 9

-9 – 4 = -13

=
9/-13

The slope is -9/13

In this case it is negative because you are dividing a positive integer by a negative integer, so the slope is negative.

If you have an even number of negative integers that you’re multiplying or dividing the answer is positive, if you have an odd number of negative integers that you’re multiplying or dividing, then the answer will be negative.

-2 x -2 x -2 = -8
-2 x -2 x 2 = 8

-24 / -12 / -2 = -1
-24 / -12 / 2 = 1

I hope this is helpful to you.