gibsonj338

To solve this problem, first find the slope of the first two points.

\(m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\)

\(m=slope\)

\({y}_{2}=\)y-point in second point

\({y}_{1}=\)y-point in first point

\({x}_{2}=\)x-point in second point

\({x}_{1}=\)x-point in first point

\({y}_{2}=3\)

\({y}_{1}=4\)

\({x}_{2}=6\)

\({x}_{1}=2\)

\(m=\frac{3-4}{6-2}\)

\(m=\frac{-1}{6-2}\)

\(m=\frac{-1}{4}\)

\(m=-\frac{1}{4}\)

Next, do the same for the second and third points and solve for c.

\(m=-\frac{1}{4}\)

\({y}_{2}=c\)

\({y}_{1}=3\)

\({x}_{2}=-5\)

\({x}_{1}=6\)

\(-\frac{1}{4}=\frac{c-3}{-5-6}\)

\(-\frac{1}{4}=\frac{c-3}{-11}\)

\(-\frac{1}{4}\times-11=\frac{c-3}{-11}\times-11\)

\(\frac{11}{4}=\frac{c-3}{-11}\times-11\)

\(\frac{11}{4}=c-3\)

\(\frac{11}{4}+3=c-3+3\)

\(\frac{11}{4}+\frac{12}{4}=c-3+3\)

\(\frac{11+12}{4}=c-3+3\)

\(\frac{23}{4}=c-3+3\)

\(5.75=c-3+3\)

\(5.75=c-0\)

\(5.75=c\)

\(c=5.75\)

To answer the first queston, first factor the polynomial function.

\(y={x}^{3}-{3x}^{2}+16x-48\)

\(y=({x}^{3}-{3x}^{2})+(16x-48)\)

\(y={x}^{2}(x-3)+(16x-48)\)

\(y={x}^{2}(x-3)+16(x-3)\)

\(y=({x}^{2}+16)\times(x-3)\)

\(y=({x}^{2}+{4}^{2})\times(x-3)\)

\(y=(x-4)\times(x+4)\times(x-3)\)

Now set each x value to zero and solve

\(x-4=0\)

\(x-4+4=0+4\)

\(x-0=0+4\)

\(x=0+4\)

\(x=4\)

\(x+4=0\)

\(x+4-4=0-4\)

\(x+0=0-4\)

\(x=0-4\)

\(x=-4\)

\(x-3=0\)

\(x-3+3=0+3\)

\(x-0=0+3\)

\(x=0+3\)

\(x=3\)

Now plug each answer to the original polynomial function to see if the answers will fit in the original polynomial function.

\(y={x}^{3}-{3x}^{2}+16x-48\)

\(0={x}^{3}-{3x}^{2}+16x-48\)

\(0={4}^{3}-3({4}^{2})+16(4)-48\)

\(0=64-3({4}^{2})+16(4)-48\)

\(0=64-3(16)+16(4)-48\)

\(0=64-48+16(4)-48\)

\(0=64-48+64-48\)

\(0=16+64-48\)

\(0=80-48\)

\(0≠32\)

\(y={x}^{3}-{3x}^{2}+16x-48\)

\(0={x}^{3}-{3x}^{2}+16x-48\)

\(0={(-4)}^{3}-{3(-4)}^{2}+16(-4)-48\)

\(0=-64-{3(-4)}^{2}+16(-4)-48\)

\(0=-64-3(16)+16(-4)-48\)

\(0=-64-48+16(-4)-48\)

\(0=-64-48+(-64)-48\)

\(0=-112+(-64)-48\)

\(0=-176-48\)

\(0≠-224\)

\(y={x}^{3}-{3x}^{2}+16x-48\)

\(0={x}^{3}-{3x}^{2}+16x-48\)

\(0={3}^{3}-{3(3}^{2})+16(3)-48\)

\(0=27-{3(3}^{2})+16(3)-48\)

\(0=27-3(9)+16(3)-48\)

\(0=27-27+16(3)-48\)

\(0=27-27+48-48\)

\(0=0+48-48\)

\(0=48-48\)

\(0=0\)

The only real answer is x=3.  Becasue x=4 and x=-4 are not real answers, that means they are imaginary answers.  This means that B is the correct answer.

I know how to solve the second question but, because I do not know how to type it out on this forum, I will leave that to someone who does know how to answer the question and how to type it out on this forum.

\({r}^{4}=0.7083333\)

\(\sqrt[4]{{r}^{4}}=\sqrt[4]{0.7083333}\)

\(r=\sqrt[4]{0.7083333}\)

\(r=0.9174014343389823...\)

\(r≈0.92\)

To find an equation that has the three solutions of 2, 4, and 5, first set each solution equal to x.

\(x=2\)

\(x=4\)

\(x=5\)

Next, figure out what number goes with each of the three solutions that makes x equal to zero on the right side.  For x=2, the number will be -2; for x=4, the number is -4; for x=5, the number will be -5. What you do on the right side, you do to the left side as well.

\(x=2\)

\(x-2=2-2\)

\(x-2=0\)

\(x=4\)

\(x-4=4-4\)

\(x-4=0\)

\(x=5\)

\(x-5=5-5\)

\(x-5=0\)

Next, combine all three equations using the zero product.

\((x-2)\times(x-4)\times(x-5)=0\)

Next, expand the left side.

\((x-2)\times(x-4)\times(x-5)=0\)

\(({x}^{2}-4x-2x+8)\times(x-5)=0\)

\(({x}^{2}-6x+8)\times(x-5)=0\)

\({x}^{3}-5{x}^{2}-6{x}^{2}+30x+8x-40=0\)

\({x}^{3}-11{x}^{2}+38x-40=0\)

Next, replace 0 on the right side with y and then switch the equation around so that the y is on the left side.

\({x}^{3}-11{x}^{2}+38x-40=y\)

\(y={x}^{3}-11{x}^{2}+38x-40\)

The negetave radical distance means to go the distance in the oppisite direction.  It is very easy to visualize what this means.

Polar coordiantes can be represented as (r, θ) where r equals the radius and θ equals the angle in degrees or in radians.  To convert the cartesian coordinate (2,-2) to polar coordinate, first figure out what r is. To find out what r is, use the formula known as pythagoras theorem: \({r}^{2}={x}^{2}+{y}^{2}\) where x is the x-coordinate and y is the y-coordinate.

\({r}^{2}={x}^{2}+{y}^{2}\)

\({r}^{2}={2}^{2}+{(-2)}^{2}\)

\({r}^{2}=4+{(-2)}^{2}\)

\({r}^{2}=4+4\)

\({r}^{2}=8\)

\(\sqrt{{r}^{2}}=\sqrt{8}\)

\(r=\sqrt{8}\)

\(r=2\sqrt{2}\)

Now figure out what θ is.  To figure out what θ is, use the formula known as tangent function:

\(tan(\Theta)=\frac{x}{y}\).

\(tan(\Theta)=\frac{x}{y}\)

\(tan(\Theta)=\frac{2}{-2}\)

\(tan(\Theta)=-\frac{2}{2}\)

\(tan(\Theta)=-1\)

\({tan}^{-1}(tan(\Theta))={tan}^{-1}(-1)\)

\(\Theta ={tan}^{-1}(-1)\)

\(\Theta =-45°\)or \(\Theta =-\frac{\pi}{4}\)

Because the question asks to be within the 0° ≤ θ < 360° parameter, ignore the radian answer above.  Since the degree answer is not within the 0° ≤ θ < 360° parameter, you need to change the answer to an equilivent answer that fits the 0° ≤ θ < 360°.  To do that add 360° to the degree answer.

\(\Theta=-45°+360°\)

\(\Theta=315°\)

Now put r and θ in polar cordinate form.

\((r,\Theta)\)

\((2\sqrt{2}, 315°)\)

To find another coordinate in polar form that is the same as the polar coordinate above that fits the 0° ≤ θ < 360°, first subtract 180° from 315°.

\(\Theta=315°-180°\)

\(\Theta=135°\)

Second, change \(2\sqrt{2}\) to \(-2\sqrt{2}\).

Now put r and θ in polar cordinate form.

\((r,\Theta)\)

\((-2\sqrt{2},135°)\)

I assume you mean \(\frac{8}{3}-\sqrt{6}\)

\(\frac{8}{3}-\sqrt{6}\)

\(\frac{8}{3}-\frac{3\sqrt{6}}{3}\)

Exact answer:

\(\frac{8-3\times\sqrt{6}}{3}\)

Approximate answer:

\(\frac{8-3\times\sqrt{6}}{3}\)

\(\frac{8-3\times2.4494897427831781}{3}\)

\(\frac{8-7.3484692283495343}{3}\)

\(\frac{0.6515307716504657}{3}\)

\(0.2171769238834885667\)

To find out how many hours Kylie needs to work to earn $357, you need to first find out how much she earns in one hour. Since she made $68 for 4 hours of work, to find out how much she earns in one hour, take $68 and divide that amount by 4 and when you do that you get $17 for one hour of work.  To find the number of hours she needs to work to earn $357, take $357 and divide it by how much she earns in one hour which is $17 and when you do that you get 21 hours.  It will take 21 hours for Kylie to earn $357.

I do not understand how you got to \(sin(\Theta)=1.33sin(\Theta)/(-0.1135188)\); however, this is how I would sove this equation.

\(1.33\times sin(25) = 1.5\times sin(\Theta)\)

\(1.33\times -0.132351750098 ≈ 1.5\times sin(\Theta)\)

\(-0.17602782763034 ≈ 1.5\times sin(\Theta)\)

\(\frac{-0.17602782763034}{1.5} ≈ \frac{1.5\times sin(\Theta)}{1.5}\)

\(\frac{-0.17602782763034}{1.5} ≈ \frac{1\times sin(\Theta)}{1}\)

\(\frac{-0.17602782763034}{1.5} ≈ 1\times sin(\Theta)\)

\(\frac{-0.17602782763034}{1.5} ≈ sin(\Theta)\)

\(-0.1173518850868933 ≈ sin(\Theta)\)

\({sin}^{-1}(-0.1173518850868933) ≈ {sin}^{-1}[sin(\Theta)]\)

\(-0.11762291934 ≈\Theta\)

\(\Theta≈-0.11762291934\)

I do not know how to solve the second problem; however, I can help solve the first problem.  To solve the first problem, use Heron's Formula.

\(A=\sqrt{\frac{a+b+c}{2}\times(\frac{a+b+c}{2}-a)\times(\frac{a+b+c}{2}-b)\times(\frac{a+b+c}{2}-c)}\)

A = Area

a = side a or \(\frac{4}{3}\)

b = side b or \(2\)

c = side c or \(\frac{8}{3}\)

\(A=\sqrt{\frac{\frac{4}{3}+2+\frac{8}{3}}{2}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{\frac{\frac{4}{3}+\frac{6}{3}+\frac{8}{3}}{2}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{\frac{\frac{10}{3}+\frac{8}{3}}{2}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{\frac{\frac{18}{3}}{2}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{\frac{\frac{18}{3}}{2}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{3\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{3\times(\frac{\frac{4}{3}+\frac{6}{3}+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{3\times(\frac{\frac{10}{3}+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{3\times(\frac{\frac{18}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{3\times(\frac{6}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{3\times(\frac{18}{6}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{3\times(\frac{18}{6}-\frac{8}{6})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{3\times(\frac{10}{6})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{3\times(\frac{5}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{3\times\frac{5}{3}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{\frac{3}{1}\times\frac{5}{3}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{\frac{15}{3}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{5\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{5\times(\frac{\frac{4}{3}+\frac{6}{3}+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{5\times(\frac{\frac{10}{3}+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{5\times(\frac{\frac{18}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{5\times(\frac{6}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{5\times(3-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{5\times(1)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{5\times1\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{5\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{5\times(\frac{\frac{4}{3}+\frac{6}{3}+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{5\times(\frac{\frac{10}{3}+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{5\times(\frac{\frac{18}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{5\times(\frac{6}{2}-\frac{8}{3})}\)

\(A=\sqrt{5\times(3-\frac{8}{3})}\)

\(A=\sqrt{5\times(\frac{9}{3}-\frac{8}{3})}\)

\(A=\sqrt{5\times(\frac{1}{3})}\)

\(A=\sqrt{5\times\frac{1}{3}}\)

\(A=\sqrt{\frac{5}{1}\times\frac{1}{3}}\)

\(A=\sqrt{\frac{5}{3}}\)