I guess the baby has to press on 1 2 in that sequence.
So the probability of this is just
\(p=\left(\dfrac{1}{11}\right)^3 = \dfrac{1}{1331}\)
really you should have been able to get this one...
You've got 8 possible lengths and 8 possible widths.
by the congruence restriction a x b is the same as b x a
there are 8 squares
then there are 8x7 possible rectangles but we only get half of these due to the congruence restriction, so there are 28 rectangles.
28+8 = 36
\(\text{We can let $a=\cos(x),~b=\sin(x)$}\\ r (x)= \cos(x)+\sin(x)\\ \dfrac{dr}{dx} = -\sin(x)+\cos(x)\\ \dfrac{dr}{dx} =0 \Rightarrow x = \dfrac{\pi}{4}, ~\dfrac{5\pi}{4}\\ \dfrac{d^2r}{dx^2} = -\cos(x)-\sin(x)\\ \left. \dfrac{d^2r}{dx^2}\right|_{x=\frac \pi 4} <0 \text{ so this is the maxima}\\ \left. \dfrac{d^2r}{dx^2}\right|_{x=\frac {5\pi} 4} >0 \text{ so this is the minima}\\ r\left(\frac \pi 4\right) = \sqrt{2}\\ r\left(\frac{5\pi}{4}\right) = -\sqrt{2}\\ \text{and we get that $a+b \in [-\sqrt{2},\sqrt{2}]$} \)
1 letter words
1,2 - 2 words
2 letter words
10-27 - 18 words
3 letter words
100-270 - 171 words
4 letter words
1000-2707 - 1708 words
1708+171+18+2 = 1899
8 1 digit numbers, leaves 7 each of the digits 1-8
81-87 uses all the 8s and leaves 6 each of 1-7
71-76 uses all the 7s and leaves 5 each of 1-6
61-65 leaves 4 each of 1-5
51-54
41-43
31-32
21
8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = (8)(9)/2 = 36
when k=6 the solutions for x,y,z all have 0 in the denominator which is a nono
\(\text{textbook partial fractions problem}\\ x+7 = A(x+1) + B(x-2)\\ x+7 = (A+B)x + (A-2B)\\ A+B=1\\ B=1-A\\~\\ A-2B=7\\ A - 2(1-A) = 7\\ 3A - 2 = 7\\ A=3\\ B=-2\\~\\ \dfrac{x+7}{x^2-x-2}= \dfrac{3}{x-2}- \dfrac{2}{x+1}\)
\(f(5) = 3f(3)-2f(4)\\ f(3) = 3f(1)-2f(2) = -3 - 2(3) = -9\\~\\ f(4) = 3f(2)-2f(3) = 3(3) - 2(-9) = 27\\~\\ f(5) = 3(-9) - 2(27) = -81\)
list them
I maintain that 15 and 36 are the correct answers.
1)
\(\text{The remainder can be any degree $0-4$}\)
2)
\(\text{by the polynomial remainder theorem}\\ p(-1)=5\\ p(-5)=-7\\~\\ p(x)\pmod{(x+1)(x+5)} = ax+b\\ p(x) = q(x)(x+1)(x+5) + (ax+b)\\ p(-1) = 5 = 0+(-a+b)\\ p(-5) = -7 = 0 + (-5a+b)\\ 4a= 12\\ a=3\\ b=8\\~\\ p(x) \pmod{(x+1)(x+5)} = 3x+8\)
+6252 