amygdaleon305

We have, 

\(sin(x+{\pi \over 2})-cos(x-\pi)=1\)

⇒\(cosx-cos(x-\pi)=1\)

Now, we know the formula   \(cosx-cosy=-2sin{x+y\over 2}sin{x-y \over 2}\)

⇒\(-2sin{x+x-\pi\over 2}sin{x-x+\pi\over 2}=1\)

⇒\(-2sin({x-{\pi\over 2}})sin{\pi\over 2}=1\)

⇒\(2sin({\pi\over 2}-x)sin{\pi\over 2}=1\)

⇒\(2cosxsin{\pi\over 2}=1\)

⇒\(cosx=1/2\)

⇒\(x={\pi\over 3}\)

~Hope this helps :)

Yeah you're right, in that case lets eliminate one possibility and make it 494 possible polynomials.

It is given that  \(-1 ≤ x≤ 1\) 

Case 1: when \(x=-1\)

\(\sqrt{58-42x}+\sqrt{149-140\sqrt{1-x^2}} = \sqrt{100}+\sqrt{149}\)

                                                                 \(=10+7\)

                                                                 \(=17\)

Case 2: when \(x=0\)

\(\sqrt{58-42x}+\sqrt{149-140\sqrt{1-x^2}}=\sqrt{58}+\sqrt9 \)

                                                                 \(=7.6+3\)

                                                                 \(=10.6\)

Case 3: when \(x=1\)

\(\sqrt{58-42x}+\sqrt{149-140\sqrt{1-x^2}}=\sqrt{16}+\sqrt{149}\)

                                                                 \(=4+7\)

                                                                 \(=11\)

Comparing cases 1, 2 and 3

Minimum value =10.6 

∴ Minimum possible value of given expression is 10.6 

~Hope you got it:) 

Consider a third-degree polynomial \(f(x)\) such that

        \(f(x) =ax^3+bx^2+cx+d\)

⇒    \(f(1) = a+b+c+d\)

Now, since  \(f(1)=9\)

∴ \(a+b+c+d=9\)

So, you're looking for the number of ways to write 9 as the sum of four non-negative integers.

Here I'm applying "sticks and stones" method as you suggested.

⇒No. of solutions \(=\binom{n+r-1}{r-1}\)

Here, \(n=9\)  and  \(r=4\)

Total no. of possible solutions  \(=\binom{9+4-1}{4}\)

                                                 \(=\binom{12}{4}\)

                                                 \(={12! \over (12-4)!4!}\)

                                                 \(={12! \over 8!4!}\)

                                                 \(=495\)

∴ There are 495 possible third-degree polynomials that fulfills the given condition. 

Phew! This question blew my mind! 

P.S. Suggestions and corrections are welcome.

Let's assume Will and Elvin had \(x\) no. of bookmarks at first. 

After Will gave away 29 bookmarks and Elvin gave away 48 bookmarks, Will had 2 times as many bookmarks as Elvin.

So, according to question

\(x-29=2(x-48)\)

\(x-29=2x-96\)

⇒\(x=67\)

∴ Will and Elvin had 67 bookmarks each, at first. 

~Hope this helps :) 

Given equation is 

\(9y^2={3^5 \over 27^3}\)

⇒\((3y)^2={3^5 \over 3^9}\)

⇒\((3y)^2=3^{-4}\)

⇒\(y^2=3^{-6}\)

⇒\(y^2=(3^{-3})^2\)

⇒\(y=3^{-3}\)

⇒\(y={1 \over 27}\)

∴ Value of y is \({1 \over 27}\).

Let the radius of circle A be x, of B be y and of C be z. 

Now, consider the following diagram- 

From figure, 

\(x+y=14\)               ...(1)

\(y+z=18\)               ...(2)

\(z+x=18\)               ...(3)

Adding eq.(1), (2) and (3)

⇒\(2x+2y+2z=14+18+18\)

⇒\(2(x+y+z)=50\)

⇒\(x+y+z=25\)                   ...(4)

Subtracting eq.(1) from (4), 

⇒\(z=11\)

Subtracting eq.(2) from (4), 

⇒\(x=7\)

Subtracting eq.(1) from (4), 

⇒\(y=7\)

∴ Radius of circle A is 7 units, B is 7 units and C is 11 units.

~Roll the credits.

△BCD is -

By Pythagoras' theorem, 

\(CD^2=BC^2-BD^2\)

          \(=64-16\)

          \(=48 \)

\(CD=4{\sqrt{3}}\)

⇒\(sin B={CD \over BC}\)

              \(={4{\sqrt3} \over 8}\)

              \(={{\sqrt 3}\over 2}\)

∴ The value of sin B is \({{\sqrt 3}\over 2}\).

Thanks a ton for this explanation :D

Thank you so much! Melody