amygdaleon305

Its okay, its a really hard question... it may take some time to get solved 

Has anyone had any luck with this question so far? I tried everything I could, but it only seem to get more confusing 

Yeah that's very good but then in order to find the value of n we would've to find the value of c/a which are the coefficients in f(x). 

Anyway, thanks a lot catmg! :D Let me see what can be done further

Melody,

From what I understand it says that f(x) is a polynomial of degree atmost 7 right? then it would have atmost 7 roots.

Let's then consider its roots are a,b,c,d,e,f,g then, sum of pairwise product of all roots means ab + bc + cd + de + ef + fg + ag + bg + cg +...+ eg + af + bf +.... so on = n you get the point right?

I think then we would've to find that sum to find the value of n. 

Another thing, its also given that f(x) leaves remainder -1 and 1 on division with (x-1)⁴ and (x+1)⁴ I don't know what to deduce from that...

Oh I see it now okay! Thanks a lot :)

But as per the solution, at one point you've seem to take n instead of n₁ and n₂ what about that? That's why I presumed the solution might be holding for  n₁ = n₂

Yes indeed option d is too obvious, option c would be the right one since you arrived at the answer by only taking n₁ = n₂ (= n) commonly right? 

Thank you so much!

"He charges 1 euro cent for the first meter, but then the price is to be increased by 50% from meter to meter."

Let's take this into account as an GP such that

       GP = 1, 1.5, 2.25,... and so on

⇒ a = 1     and      r = 1.5 

Length of the well is 37 m i.e. the no. of terms in the GP is 37. 

Now, in order to calculate the total amount, we find the sum of 37 terms in the GP

         ∵ \({S}_{n}={a(r^n-1)\over r-1}\)

        ⇒ \({S}_{37} = {1((1.5)^{37}-1)\over 1.5-1} ≈ 6552491.2\)

Thus he has to pay 6,552,491.2 euro cents. 

In order to find the no. of possible sequences, you find the no. of ways 1, 2, 2, 3, 3 and 3 can be arranged. 

So, no. of digits = 6

      no. of 2's = 2

      no. of 3's = 3

Possible no. of rolls \(={6!\over 2!×3!}=60\)

Hope you got it. 

Is it okay if I answer? 

Spikers = 5, Setters = 4 and 1 either way.

Each 2-person team should have 1 setter and 1 spiker. 

Case 1 : If spikers = 6 and setters = 4                 Case 2 : If spikers = 5 and setters = 5

Then no. of ways \(={^6} C_{1}×{^4}C_{1}\)                         Then no. of ways \(={^5} C_{1}×{^5}C_{1}\)

In order to get total no. of ways, add the respective values you get in both cases. 

\(3y=2x^2+16x+28\)

      \(=2(x^2+8x+16)-2\)

\(y={2\over 3}(x+4)^2-{2\over 3}\)

This equation is similar to  \(y=a(x-h)^2+k\)  in which (h,k) is the vertex 

⇒ (m,n) = (-4, -2/3)

 \(m+n=-4-{2\over 3} = -{14\over 3}\) 

\(x^2+13x+30 = x^2+3x+10x+30\)

                           \(=x(x+3)+10(x+3)\)

                           \(=(x+3)(x+10)\)

⇒ a = 3   and  b = 10 

\(x^2+5x-50 = x^2+10x-5x-50\)

                         \(=x(x+10)- 5(x+10)\)

                         \(=(x+10)(x-5)\)

⇒ c = 5

\(a+b+c = 3+10+5 = 18\)

So far, you're doing really well..

After \(y = {log128\over log8}\)  you don't really need a calculator to solve this 

We can write this as

   \(y= {log 2^7\over log 2^3}\)  right?

then \(y= {7log2\over 3log2}\)

     ⇒\(y={7\over 3}\)

     \({log}_{8}x = {7\over 3}\)

Hope you got it :)

\((3x^2-5x+4)(7-x) = -3x^3+26x^2-39x+28\)

a = -3, b = 26, c = -39 and d = 28

8a + 4b + 2c + d = 30

No. of black cards = No. of red cards = 26

4 black cards can be drawn in = \(^{26}C_4\) ways

1 red card card can be drawn in = \(^{26}C_1\) ways 

No. of ways  = \(^{26}C_4 × {^{26}C_1}\)

It would be easier for us if you'd enclose a diagram with question especially geometry. Anyway, 

2) ∠BDE + ∠CBD  + ∠CED + ∠BCE = 360 

    ∠BDE + 31 + 47 + 85 = 360 

     ∠BDE = 360 - 163 = 197°

Digits = 1, 2, 3, 4, 5, 6, 7, 8 and can be repeated. 

1's digit has = 4 possibilities

Rest all digits have 8 possibilities 

No. of 5-digit numbers = 8 × 8 × 8 × 8 × 4 = 16384