Rom

\(x(x+1)=272 \\ x^2 + x =272 \\ (x+1/2)^2 -\dfrac 1 4 = 272 \\ (x+1/2)^2 = \dfrac {1089}{4} \\ x+\dfrac 1 2 = \sqrt{\dfrac {1089}{4}} = \dfrac {33}{2} \\ x=\dfrac {32}{2} = 16 \\ \mbox{the two numbers are }16 \mbox{ and } 17\)

\(-16 \mbox{ and } -17 \mbox{ work as well.}\)

not seeing the "as shown" bit but that circle has a radius of 4 so the square in question is probably 8x8

It depends on whether you start with the whole pizza or just half of it.

If you start with half of it and slice it into slices with 30º angles then yes 6 slices is correct.

If you start with a whole pizza and slice half of it as above you'll end up with 7 slices.  One of which is half the pizza.

the successive differences are

2 1/2, 5, 10

and you can see that these are doubling each time

\(\mbox{A recursive rule would be} \\ a[n] = a[n-1] + \left(\dfrac 5 2\right)\times 2^{n-1} ; a[0]=3\)

\(\mbox{An explicit rule would be} \\ a[n] = \dfrac {1+5 \times 2^n}{2}\)

P(A+B) is NEVER greater than 1

\(F_{friction}=m g \mu_f = (3)(9.8)(0.28) = 8.232 N\\ 12 ft = 3.6576 m \\ Work = 8.232 N \times 3.6576m = 30.11 Nm = 30.11 J\)

\(\dfrac 6 {42} = \dfrac 1 7\)

\(0.5 x \ln(x) -1.5 x = 0 \\ 0.5 x \ln(x) = 1.5 x \\ x \ln(x) = 3 x \\ \mbox{Since }\ln(0) \mbox{ is not defined }x \neq 0 \mbox{ and we can divide both sides by }x \\ \ln(x) = 3 \\ x = e^3\)

https://www.youtube.com/watch?v=XVSRm80WzZk

\(\mbox{If }x=2 \mbox{ is a root of }x^3-7x+6 \mbox{ then } (x-2) \mbox{ divides } x^3-7x+6\)

\(x^3-7x+6 = x^2(x-2) + (2x^2-7x+6) \\ (2x^2-7x+6)=2x(x-2) +(-3x + 6) \\ (-3x+6) = -3(x-2) \\ x^3-7x+6 = (x-2)(x^2+2x-3)\)

\(\mbox{To find the rest of the zeros we have to factor }x^2+2x-3 \\ \mbox{and this is easily done resulting in } \\ x^2+2x-3 = (x+3)(x-1) \\ \mbox{and thus the other zeros are } x=-3, x=1\)