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antoniya [11.8K]
2 years ago
8

PLEASE ANSWER NUMBER 2 3 AND 4 THANK YOU

Mathematics
1 answer:
Darina [25.2K]2 years ago
5 0
I can’t really see it. take another pic please
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The values of sin θ and cos θ represent the legs of a right triangle with a hypotenuse of 1; therefore, solving for cos θ finds the unknown leg, and then all other trigonometric values can be found.

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I really need help please!
miskamm [114]
Simple...

you have: \frac{ 3^{-4}* 2^{3}* 3^{2}   }{ 2^{4}* 3^{-3}  }

Multiply it out-->>

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3 years ago
Please please please need help on 3 and 4 I’m really lost on them
Alex73 [517]

Answer:

1) Option c) is correct ie., 5 real and o non-real

2) Option b) is correct ie., (4, \frac{1}{2}, \frac{1}{2}, 2,2)

Step-by-step explanation:

Given polynomial function is f(x)=x^5-3x^3-2

To find zeros equate f(x) to zero ie.,  f(x)=0

x^5-3x^3-2=0

By synthetic division

     |   1     0    -3      0      -2

-1  |   0    -1      1      2      2

     |_________________

        1      -1      -2     2      0

Therefore x=-1 is a zero

x^3-x^2-2x+2=0

      |  1    -1     -2    2

1      |  0     1     0     2

      |___________________

          1     0   -2   0

x=1 is the zero

x^2 -2 =0

x=\pm\sqrt{2}

x=\sqrt{2} and

x=\sqrt{-2}

Option c) is correct ie., 5 real and o non-real

2) Given polynomial function is f(x)=(x-4)(2x-1)^2(x-2)^2

To find zeros equate f(x) to zero ie.,  f(x)=0

                                                            (x-4)(2x-1)^2(x-2)^2=0

(x-4)=0 (or) (2x-1)^2=0 or (x-2)^2=0

Therefore x=4, x=\frac{1}{2} of multiplicity of 2 and x=2 multiplicity of 2

Option b) is correct ie., (4, \frac{1}{2}, \frac{1}{2}, 2,2)

5 0
3 years ago
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