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My name is Ann [436]
4 years ago
5

Using the given zero, find one other zero of f(x). Explain the process you used to find your solution.

Mathematics
1 answer:
umka21 [38]4 years ago
7 0
Complex zeroes always occurs as conjugates.
For z = a + b i conjugate is: a - b i
Another zero is : 2 + 3 i.
Verification:
2 + 3 i + 3 - 3 i = - b/a
- b = 4,  a = 1
( 2 + 3 i ) ( 2 - 3 i ) = c / a
4 - 9 i² = c / a
4 + 9 = c / a
c = 13
( x^4 - 4 x³ + 14 x² - 4 x + 13 ) : ( x² - 4 x + 13 ) = x² + 1
x² + 1 = 0
x² = -1,  x = i,  x = -i
The zeroes are: - i , i , 2 + 3 i, 2 - 3 i.
Answer: 
One another zero of f ( x ) is 2 + 3 i.
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4 0
3 years ago
Donna wants to make trail mix made up of almonds, walnuts, raisins. she wants to mix one part almonds, two part walnuts, and thr
kicyunya [14]
Almons:walnuts:rasins=1:2:3


x=pounds of walnuts
y=pounds of almonds
z=pounds or rasins
2x=y
3x=z
subsitute


12x+9y+5z=15
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8 0
3 years ago
Find the volume of the largest circular cone that can<br> beinscribed in a shpere of radius 3.
Aleksandr [31]

Answer:

V =\dfrac{32}{3}\pi

Step-by-step explanation:

given,

radius of sphere = 3

volume of cone:

V = \dfrac{1}{3}\pi r^2h

r is the radius of circular base

h is the height of the cone

here r = x and h = 3 + y

now, volume in term of x and y

V = \dfrac{1}{3}\pi x^2(3+y)

Applying Pythagoras theorem

x² + y² = 3²

x = \sqrt{9-y^2}

V = \dfrac{1}{3}\pi ( \sqrt{9-y^2})^2(3+y)

V = \dfrac{1}{3}\pi ( 9-y^2)(3+y)

V = \dfrac{1}{3}\pi (27 + 9 y - 3 y^2-y^3)

differentiating both side

\dfrac{dV}{dy} =\dfrac{1}{3}\pi ( 9-6y- 3y^2)

for maxima  \dfrac{dV}{dy} = 0

\pi ( 3-2 y - y^2)=0

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(y+3)(y-1)=0

 y = 1,-3

y cannot be negative so, volume at y = 1

V = \dfrac{1}{3}\pi (27 + 9 (1)- 3(1)^2-(1)^3)

V =\dfrac{32}{3}\pi

Hence, the largest cone which can be inscribed in the spheres of the radius 3 has volume  V =\dfrac{32}{3}\pi

5 0
3 years ago
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Step-by-step explanation:

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