Vinculum

a)  To find the number of boxes in any row...we can use

an  =  63 - 7 (n - 1)      where  n  is an integer from 1 to 6 inclusive

So...in the 6th row, we have

a6  =  63 - 7(6 - 1)  =  28 boxes

b)  To find the total number of boxes, we can use

Sum  = 

[ Number of boxes in the first row + Number of boxes in last row ] * [ No. of rows / 2 ]

So we have

[ 63 + 28 ] * [6/2]   =

273 boxes in total

Used answer by CPhill for framework: https://web2.0calc.com/questions/anyone_11

-Vinculum

Question 4:

-Vinculum

Used: https://socratic.org/answers/580659

All credits to the person that answered it there

Question 3.

Used: https://socratic.org/answers/580941

-Vinculum 

G'day mate, 

"y" in this case is "a", I accidentally used "a" throughout, sorry!!!!!

\(\begin{bmatrix}\frac{1}{x}+\frac{1}{a}=\frac{1}{3}\\ xa+x+a=4\end{bmatrix}\)

\(\mathrm{Plug\:the\:solutions\:}x=-\frac{3a}{3-a}\mathrm{\:into\:}xa+x+a=4\)

We get a = \(a=\frac{1}{2}+i\frac{\sqrt{11}}{2} .and. a=\frac{1}{2}-i\frac{\sqrt{11}}{2}\)

\(\mathrm{Plug\:the\:solutions\:}a=\frac{1}{2}-i\frac{\sqrt{11}}{2},\:a=\frac{1}{2}+i\frac{\sqrt{11}}{2}\mathrm{\:into\:}\frac{1}{x}+\frac{1}{a}=\frac{1}{3}\)

That gives \(x=\frac{1}{2}+\frac{\sqrt{11}}{2}i and x=\frac{1}{2}-\frac{\sqrt{11}}{2}i\)

So now we have our values for (x, y): \(\begin{pmatrix}x=\frac{1}{2}+\frac{\sqrt{11}}{2}i,\:&y=\frac{1}{2}-i\frac{\sqrt{11}}{2}\\ x=\frac{1}{2}-\frac{\sqrt{11}}{2}i,\:&y=\frac{1}{2}+i\frac{\sqrt{11}}{2}\end{pmatrix}\)

Plugging that in: \(x^2\cdot y + x\cdot y^2\):

\(\left(\frac{1}{2}+\frac{\sqrt{11}}{2}i\right)^2\left(\frac{1}{2}-\frac{\sqrt{11}}{2}i\right)+\left(\frac{1}{2}+\frac{\sqrt{11}}{2}i\right)\left(\frac{1}{2}-\frac{\sqrt{11}}{2}i\right)^2=3\)

\(\left(\frac{1}{2}-\frac{\sqrt{11}}{2}i\right)^2\left(\frac{1}{2}+\frac{\sqrt{11}}{2}i\right)+\left(\frac{1}{2}-\frac{\sqrt{11}}{2}i\right)\left(\frac{1}{2}+\frac{\sqrt{11}}{2}i\right)^2=3\)

So the answer is \(3.\)

-Vinculum 

The constant term of \(\left(x+x^{-\frac{3}{2}}\right)^{15}\) when expanded is \(5005\).

When the above expression is expanded, we get 

Applying Binomial Therom:

\(\sum _{i=0}^{15}\binom{15}{i}x^{\left(15-i\right)}\left(x^{-\frac{3}{2}}\right)^i\)

\(x^{15}+15x^{\frac{25}{2}}+105x^{10}+455x^{\frac{15}{2}}+1365x^5+3003x^{\frac{5}{2}}+5005+\frac{6435}{x^{\frac{5}{2}}}+\frac{6435}{x^5}+\frac{5005}{x^{\frac{15}{2}}}+\frac{3003}{x^{10}}+\frac{1365}{x^{\frac{25}{2}}}+\frac{455}{x^{15}}+\frac{105}{x^{\frac{35}{2}}}+\frac{15}{x^{20}}+\frac{1}{x^{\frac{45}{2}}}\)

The constant term as seen above is \(5005\).

-Vinculum

What happened?

The inequality will be greater than zero

 if D is less than zero.

D<0

4(4k−1)2−4.1.(15k2−2k−7)<0

Simplifying this we get

K2−6k+8<0

(k−4)(k−2)<0

∴k lies in (2,4)

In this interval, only one integer is present i.e 3

∴k=3

-Vinculum

Used as reference:

https://www.toppr.com/ask/en-us/question/the-integer-k-for-which-the-inequality-x2-24k-1x-15k2/

This is when the expression has to equal (a + 3)/0

a^3 - a  =  0   =   a(a+1)(a-1)

So it's undefined when a is -1 and 1.

-Vinculum