Member: Melody

Quote:
At least one root is between 1 and 2

According to Descartes rule of signs, we have one positive root, and either two negative roots or no negative roots

Okay Chris,
I am not familiar with Descartes rule - Yes I know I could look it up - okay I did.

Now P(1) is negative and P(2) is positive therefore there must be at least root between 1 and 2. That is obvious.

http://en.wikipedia.org/wiki/Descartes'_rule_of_signs
EXTRACT
"Positive roots[edit]
The rule states that if the terms of a single-variable polynomial with real coefficients are ordered by descending variable exponent, then the number of positive roots of the polynomial is either equal to the number of sign differences between consecutive nonzero coefficients, or is less than it by an even number. Multiple roots of the same value are counted separately.

Are 2x^2 -4 counted as consecutive since there is a zero term inbetween ? If they are then there is 1 sign change. Hence there is 1 positive root.

Could you comment please Chris?
Thankyou.

PS i still think it was a copy error.

MelodyMar 20, 2014

+118725

Guest:
What are the roots for P(x) = x^3 + 2x^2 -4

Are you sure the signs are right? The answer is not popping out at me!

Hint: You have to look fo an x value that will make P(x)=0

then look here
http://www.mathsisfun.com/algebra/polynomials-remainder-factor.html

MelodyMar 20, 2014

+118725

Guest:
2X, given sinX= 2/5 and cos X <0

sinx is pos therefore it is 1st or 2nd quad
cosx is neg therefor 2nd or 3rd quad
Therefore x must be in the second quadrant

sin ^-1(2/5) = 23.578.. (first quad)
X = 180-23.578... = 156.4218...
In general X=156.4218 + 360N Where N is in the set of integers

2x = 312.84...
2x = 312 degrees 50minutes and 37 seconds +720N Where N is in the set of integers (to the nearest second)

MelodyMar 20, 2014

+118725

bump

MelodyMar 20, 2014

+118725

luzmartinez:
how to do exponents

Try this
http://www.mathsisfun.com/algebra/exponents-logarithms.html

MelodyMar 20, 2014

+118725

HelpmePlease:
How can you tell if a graph is a linear function?

Rom is right of course
but
do you mean how do you tell if an equation will be a straight line?

It will have no squared terms
It will have no x*y terms
It will have no x or y in the denominators.
so it might look like

y=3x+4
2x+3y=7
5x/4 - 6x = 13/3
0.2y = 12- 0.8x
y=15

but it won't look like
xy=8
3/x=y
6x-x ²=4

I hope that helps.

MelodyMar 20, 2014

+118725

jmsylvia:
25.12/2pi

This is 25.12/2 *pi

or do you mean

25.12/(2*pi)

Either way you can do it on a calculator. there is one on this site. Look at the top of the page.

MelodyMar 20, 2014

+118725

Nobody Special:
In an arithmetic sequence a_8 = 9 and a_12 = 37. What is a_13?
Is there an equation or something that I'm missing here? Something to simplify it?

If anyone has any ideas, that would be awesome! ^_^

Hi Special,
I am going to look after you myself.

a_8 should be written as a ₈ I used the 'sub' button above the smilies to get this.

Just press a 'sub' put the 8 where the curser is, then move curser to the end and keep going. The 'sub' stands for subtext.
Most likely it is written as a_8 because this is the code for subtext in LaTex which is a maths program - It is probably the code in other computer languages as well.

now we have
It is an AP and you need to find a ₁₃
a ₈ = 9 and a ₁₂=37

the formula for the nth term of an AP is
T _n=a+(n-1)d SCRATCH THAT
sorry I just realized that you are using different letters from what I am used to, I had better change the formula accordingly
a_n = a₁ + (n-1)d [where d is the common difference][maybe this is the equation that you are missing? - It is not normally expressed like this, not in my experience anyway]
and you have
a ₈=a ₁+(8-1)*d ==> a ₁+7d = 9
a ₁₂=a ₁+(12-1)*d ==> a ₁+11d = 37

that is
a ₁+7d = 9 (1)
a ₁+11d = 37 (2)
You just have to solve these simultaneously
(2)-(1) gives you
4d = 28
d=7
Now substitute that back into 1 of the equations to get the value of a ₁
Then just substitute the values of a ₁ and d into the AP equation.
Then sub in n=13 and find the answer that you actually want.

Have a go if you don't understand I will help some more.

MelodyMar 20, 2014

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