Melody

I just answered this, near enough anyway.....

You got in a mess with this one EP   

But considering how may questions you answer correctly I do not think you need to be upset by a silly error here and there. :)

\(tan \theta = -\frac{3}{2}\\ \)

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Think about   \(tan \theta = +\frac{3}{2} \)    first

Draw a right angled triangle.

Mark an acute angle as theta.

opp = 3

adj=2

so hypotenuse  =     \(\sqrt{9+4}=\sqrt{13} \)

So for this triangle   \(cos\theta= \frac{2}{\sqrt{13}}=\frac{2\sqrt{13}}{13}\)

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Now, tan is negative in the 2nd and 4th quadrant

In the 2nd quad Cos is negative and in the 4th quad Cos is positive.

So

\(cos\theta=\pm\frac{2\sqrt{13}}{13}\)

Yes that is all it means :)

What is the value of   \(b+c \)   if     \(x^2+bx+c>0 \)     only when    \(x\in (-\infty, -2)\cup(3,\infty)\)?

Consider the concave up parabola   y=x^2+bx+c

\(x^2+bx+c>0 \)    will be true when y is positive, which is when the graph is above the x axis.

The roots of this graph are 

        \(x=\frac{-b\pm\sqrt{b^2-4c}}{2}\\ x=\frac{-b}{2}\pm\frac{\sqrt{b^2-4c}}{2}\\\)

the axis of symmetry of this graph will be     \(x=\frac{-b}{2}\)

No consider the domain where x is poistive.

It is less than -2 and greater then 3  SO the axis of symmetry must be half way between these points.  (-2+3)/2=-1/2

so

\(\frac{-b}{2}=\frac{-1}{2}\\ b=1\)

Also

\(\frac{-b}{2}+\frac{\sqrt{b^2-4c}}{2}=3\\ -b+\sqrt{b^2-4c}=6\\ sub\;in\;b=1\\ -1+\sqrt{1^2-4c}=6\\ \sqrt{1-4c}=7\\ 1-4c=49\\ -4c=48\\ c=-12\\ ~\\ b+c=1+-12=-11\)

Rosala, when you get the answers it would be good if you share them with us.

Make a new post that directs us to this one so that we will see it please   

i know you all might be bored by now with this but i cant help it...ive got some more PnC questions...and there are more like these so if anyone could help thatd be great!

1. Find the no. of arrangements of the letters 'abcd' in which neither a,b nor c,d come together is? 

I will assume they will be in a row.

There are 4! arrangments altogether.  4*3*2*1 = 24

In how many are they together..

I just counted

So I think that is 16

24-16=8

So that leaves 8 permutations where A is not next to B and C is not next to D

So this one agrees wit the other answerers  

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2. No. of 9 digit nos. divisible by 9 using the digits from 0 to 9 if each digit is used atmost once is K.8! , then find the K has the value equal to_______.

The digits will have to add to a multiple of 9

1,2,3,4,5,6,7,8,9 add to 45            9! =  9*8! possibilities

0,1,2,3,4,5,6,7,8 add to 36            8 possibles for the highest vaue digit (can't be zero) * 8! 

these are the only 2 possible sets of digits. 

So that is  9*8! + 8*8! = 17*8!

K=17

3. Fine the no. of 5 digit nos. such that the sum of their digits is even is.

You will need an even number of odd digits for this to work.

So All even, or 3 even or 1 even  (the other digits odd)

All even. the biggest canonot be 0.  4*5^4 = 2500

3 even and 2 odd  

e e e o o     there are 5C3 ways this can be arranged 

How many ways are there to get 3 even digits from 5 where order counts and repetition is allowed. mmm

I think that is  5^3

How many ways are there to get 2 odd digits from 5 where order counts and repetition is allowed. 

I think that is  5^2

So that seems to be     5^3*5^2*5C3 = 5^5 * 5C3 = 3125*10 = 31250 ways

But some of those have the biggest digit 0 and that is not allowed.   

How many have the first digit =0???

5^2*5^2* 4C2 = 5^4 * 4C2 = 625 * 6 = 3750

so 

31250-3750 = 27500

I think there are 27500 five digit numbers with 3 even digits and 2 odd digits

How many numbers are there with 1 even digit and 4 odd ones.

0 followed by 4 odds ... there are    5^4 = 625 (these are NOT included)

There are now 5 even ones to chose from and this can go 1st, 2nd, 3rd, 4th or 5th that is 5 places.

So that is 25 possibilities for what even one goes where.

Now there are 5^4 ways to chose the odds = 625 ways

Altogether there are  625*25 = 15625 numbers with 4 odd and 1 even digit but I have to subtract the ones that start in 0

15625 - 625 = 15000

I think there are 15000 five digit numbers with 1 even digits and 4 odd digits

So altogether that makes  2500+27500+15000 = 45000 ways.

Just like the others have said, I am not certain this is correct. 

4. Find the no. of natural nos. less than 1000 and divisble by 5 can be formed with the ten digits, each digit not occuring more than once in each number is______.

If the units is 0 then there are 9 choices for the 10s and 8 choices for the hundreds =     72 that end in zero

If the units is 5 and the tens is 0 then there are 8  choices for the hundreds =                    8 ending in 05

If the units is 5 and the tens is NOT 0 then there are 7  choices for the hundreds = 8*7 = 56  ending in not0 5

So that seems to be   72+8+56 = 136 numbers fit the criterion

Thanks Guest :)

Thanks Alan :)

\(\frac{x^2 + x + 3}{2x^2 + x - 6} \ge 0\\ \frac{x^2 + x + 3}{2x^2 + (4-3)x - 6} \ge 0\\ \frac{x^2 + x + 3}{2x^2 + 4x-3x - 6} \ge 0\\ \frac{x^2 + x + 3}{2x( x+2) - 3(x+2)} \ge 0\\ \frac{x^2 + x + 3}{(2x-3)( x+2) } \ge 0\\ \)

The denomator cannot be equal to zero so x cannot equal  3/2   or  -2

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The denominator by itself is  positive  for    \(-2

and it is negative for  \(x<-2\) and   negative for  \(x>1.5\)

Now I will look at just the numerator.  

consider \(y=x^2+x+3\)

this is a concave up parabola

The zeros will be  

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x = {-1 \pm \sqrt{1-12} \over 2}\\ \text{There are no real zeros} \)

this means that th numerator is positive for all real values of x.

Since the numerator is positive the quotient will be positive when and only when the denominator is positive.

So this expression will be greater than zero when     \(-2