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gibsonj338

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 #1
avatar+1904 
0

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

 

y=x33x2+16x48

 

y=(x33x2)+(16x48)

 

y=x2(x3)+(16x48)

 

y=x2(x3)+16(x3)

 

y=(x2+16)×(x3)

 

y=(x2+42)×(x3)

 

y=(x4)×(x+4)×(x3)

 

Now set each x value to zero and solve

 

x4=0

 

x4+4=0+4

 

x0=0+4

 

x=0+4

 

x=4

 

x+4=0

 

x+44=04

 

x+0=04

 

x=04

 

x=4

 

x3=0

 

x3+3=0+3

 

x0=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=x33x2+16x48

 

0=x33x2+16x48

 

0=433(42)+16(4)48

 

0=643(42)+16(4)48

 

0=643(16)+16(4)48

 

0=6448+16(4)48

 

0=6448+6448

 

0=16+6448

 

0=8048

 

032

 

y=x33x2+16x48

 

0=x33x2+16x48

 

0=(4)33(4)2+16(4)48

 

0=643(4)2+16(4)48

 

0=643(16)+16(4)48

 

0=6448+16(4)48

 

0=6448+(64)48

 

0=112+(64)48

 

0=17648

 

0224

 

y=x33x2+16x48

 

0=x33x2+16x48

 

0=333(32)+16(3)48

 

0=273(32)+16(3)48

 

0=273(9)+16(3)48

 

0=2727+16(3)48

 

0=2727+4848

 

0=0+4848

 

0=4848

 

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.

Aug 15, 2017
 #1
avatar+1904 
0

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: r2=x2+y2 where x is the x-coordinate and y is the y-coordinate.

 

r2=x2+y2

 

r2=22+(2)2

 

r2=4+(2)2

 

r2=4+4

 

r2=8

 

r2=8

 

r=8

 

r=22

 

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

tan(Θ)=xy.

 

tan(Θ)=xy

 

tan(Θ)=22

 

tan(Θ)=22

 

tan(Θ)=1

 

tan1(tan(Θ))=tan1(1)

 

Θ=tan1(1)

 

Θ=45°or Θ=π4

 

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.

 

Θ=45°+360°

 

Θ=315°

 

Now put r and θ in polar cordinate form.

 

(r,Θ)

 

(22,315°)

 

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°.

 

Θ=315°180°

 

Θ=135°

 

Second, change 22 to 22.

 

Now put r and θ in polar cordinate form.

 

(r,Θ)

 

(22,135°)

.
Aug 11, 2017
 #1
avatar+1904 
+1

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=a+b+c2×(a+b+c2a)×(a+b+c2b)×(a+b+c2c)

 

A = Area

a = side a or 43

b = side b or 2

c = side c or 83

 

A=43+2+832×(43+2+83243)×(43+2+8322)×(43+2+83283)

 

A=43+63+832×(43+2+83243)×(43+2+8322)×(43+2+83283)

 

A=103+832×(43+2+83243)×(43+2+8322)×(43+2+83283)

 

A=1832×(43+2+83243)×(43+2+8322)×(43+2+83283)

 

 

A=1832×(43+2+83243)×(43+2+8322)×(43+2+83283)

 

 

A=3×(43+2+83243)×(43+2+8322)×(43+2+83283)

 

A=3×(43+63+83243)×(43+2+8322)×(43+2+83283)

 

A=3×(103+83243)×(43+2+8322)×(43+2+83283)

 

A=3×(183243)×(43+2+8322)×(43+2+83283)

 

A=3×(6243)×(43+2+8322)×(43+2+83283)

 

A=3×(18643)×(43+2+8322)×(43+2+83283)

 

A=3×(18686)×(43+2+8322)×(43+2+83283)

 

A=3×(106)×(43+2+8322)×(43+2+83283)

 

A=3×(53)×(43+2+8322)×(43+2+83283)

 

A=3×53×(43+2+8322)×(43+2+83283)

 

A=31×53×(43+2+8322)×(43+2+83283)

 

A=153×(43+2+8322)×(43+2+83283)

 

A=5×(43+2+8322)×(43+2+83283)

 

A=5×(43+63+8322)×(43+2+83283)

 

A=5×(103+8322)×(43+2+83283)

 

A=5×(18322)×(43+2+83283)

 

A=5×(622)×(43+2+83283)

 

A=5×(32)×(43+2+83283)

 

A=5×(1)×(43+2+83283)

 

A=5×1×(43+2+83283)

 

A=5×(43+2+83283)

 

A=5×(43+63+83283)

 

A=5×(103+83283)

 

A=5×(183283)

 

A=5×(6283)

 

A=5×(383)

 

A=5×(9383)

 

A=5×(13)

 

A=5×13

 

A=51×13

 

A=53

.
Jun 30, 2017