Rom

\(\mbox{This problem leaves out a vital piece of information.}\\ \mbox{Are the coefficients of your polynomial restricted to the real numbers or can they be complex numbers?}\)

\(\mbox{Assuming they are restricted to the reals.}\\ \mbox{From the description you have polynomial factors }\\ (x-1), (x+1)^3, (x-\imath 3)\\ \mbox{Note the power of 3 for the }(x+1) \mbox{ factor comes from the meaning of multiplicity 3.}\\ \mbox{Now given the restriction on real coefficients all complex roots must come in conjugate pairs.} \\ \mbox{So you must have a further root of }(x+\imath 3).\\ \mbox{Thus }p(x)=(x-1)(x+1)^3(x-\imath 3)(x+\imath 3)\\ \mbox{I leave it to you to multiply all that out.}\)

\(\dfrac {-5} 4 + \dfrac 3 4 = \dfrac {-5 + 3}{4} = \dfrac {-2}{4}=-\dfrac 1 2\)

\(\mbox{There is no constant term so clearly }x=0 \mbox{ is one of the roots.}\\ \mbox{Dividing through by }x-0=x \mbox{ we get} \\ x^3-5x^2 +3x+1 = 0 \\ \mbox{Because the constant term is 1 if anything divides this polynomial it will be }(x-1) \mbox{ or }(x+1) \\ \mbox{Checking }(x-1)\mbox{ we plug 1 in and see if it works}\\ 1-5+3+1 = 0 \mbox{ so }1 \mbox{ is indeed a root. Dividing by }(x-1) \mbox{we get}\\ x^2-4 x-1\\ \)

\(\mbox{This can't be factored over rational numbers so we'll have to use the quadratic formula.}\\ r=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a} \\ a=1, b=-4, c=-1 \\ r_1=\dfrac{4+\sqrt{16+4}}{2}= (2+\sqrt{5}) \\ r_2 =\dfrac{4-\sqrt{16+4}}{2}= (2-\sqrt{5})\)

\(\mbox{So gathering all this up we have roots of } \\ x \in \{0,1,2+\sqrt{5},2-\sqrt{5}\}\)

\(q = p(r+s) \\ \mbox{divide both sides by }(r+s) \\ p = \dfrac{q}{r+s}\)

\(\mbox{for (2) you don't say what variable to solve for. I'm going to guess }x. \\ a x + z = a w + y \\ a x = a w + y - z \\ x = \dfrac {a w + y - z}{a}\)

Assuming the ground is flat everywhere the horizontal wind won't affect the time it takes for the ball to fall to the ground. The mass of the ball has no effect on the that time either.

\(h(t)=-\dfrac 1 2 g t^2 + v_0 t + h_0\\ h_0=161.1m \\ v_0 = 0 m/s \\ g=9.81 m/s^2 \\ \mbox{plug all that in and solve }h(t)=0 \mbox{ for }t \\ \mbox{You'll get two solutions. One will make sense.}\)

cuadrático?

pares?

continuo?

liso?

What you want to do here is use the fact that the surface area is 500cm² to solve for the length of a side of the cube or the radius of the sphere.

Then given those you can find the volume of each.

\(\mbox{A cube has surface area }SA=6s^2 \mbox{where }s \mbox{ is the length of a side.} \\ \mbox{It has volume }V=s^3 \\ \mbox{ }\\ \mbox{A sphere has surface area }SA=4\pi r^2 \\ \mbox{It has volume }V=\dfrac 4 3 \pi r^3\)

\(\mbox{First solve for y in terms of x using the second equation.}\\ \mbox{Then plug that expression for y into the first equation and solve for x.}\)

I've never seen this treated so I thought I'd take a deeper look.

\(c = r_c e^{\imath \theta_c} \\ c^\imath = \left( r_c e^{\imath \theta_c}\right)^\imath = r_c^\imath e^{\imath^2 \theta_c} = r_c^\imath e^{-\theta_c} \\ \mbox{Now what is }r_c^\imath? \\ r_c = e^{\ln(r_c)} \\ r_c^\imath = \left(e^{\ln(r_c)}\right)^\imath = e^{\imath \ln(r_c)} \\ \mbox{so }c^\imath = e^{\imath \ln(r_c)} e^{-\theta_c}= \\ e^{-\theta_c}\left(\cos(\ln(r_c))+\imath \sin(\ln(r_c))\right) \\ \mbox{Letting }c=\imath \\ r_c=1, \theta_c=\dfrac \pi 2 \\ \imath^\imath = e^{-\frac \pi 2} \left(\cos(\ln(1)) + \imath \sin(\ln(1))\right) = \\ e^{-\frac \pi 2}\left(\cos(0)+\imath \sin(0)\right)=e^{-\frac \pi 2} \\ \mbox{which is in agreement with Melody's answer}\)

\(\mbox{what do you know about the relationship between }f^\prime(x) \mbox{ and }(f^{-1})^\prime(x) ?\)