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Hitman42 [59]
2 years ago
13

Y=4x-2y and y=x+3 estimate the solutions to the systems of equations

Mathematics
1 answer:
polet [3.4K]2 years ago
6 0

Answer:

3y=4x

y=4x/3

4x/3=x+3

x/3=3

X=9

y=12

(9,12) this is the answer

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2 years ago
If z^2+1/z^2=11 , find the value of z^3-1/z^3,using only the positive value of z-1/z
vredina [299]
Hello,

Let's assume

t=z- \dfrac{1}{z} \\\\

t^2=z^2-2+ \dfrac{1}{z^2}= 11-2=9==\ \textgreater \ t=3\\\\

t^3=z^3-3z^2* \dfrac{1}{z} +3z* \dfrac{1}{z^2} - \dfrac{1}{z^3} \\\\
=z^3 - \dfrac{1}{z^3} -3(z-\dfrac{1}{z})\\\\
z^3 - \dfrac{1}{z^3}=t^3+3t=3^3+9=27+9=36



7 0
3 years ago
Write a polynomial function of minimum degree in standard form with real coefficients whose zeros and their multiplicities inclu
Softa [21]

<u>Answer-</u>

<em>The polynomial function is,</em>

y=x^4-10x^3+38x^2-64x+40

<u>Solution-</u>

The zeros of the polynomial are 2 and (3+i). Root 2 has multiplicity of 2 and (3+i) has multiplicity of 1

The general form of the equation will be,

\Rightarrow y=(x-(2))^2(x-(3+i))(x-(3-i))   ( ∵ (3-i) is the conjugate of (3+i) )

\Rightarrow y=(x-2)^2(x-3-i)(x-3+i)

\Rightarrow y=(x^2-4x+4)((x-3)-i)((x-3)+i)

\Rightarrow y=(x^2-4x+4)((x-3)^2-i^2)

\Rightarrow y=(x^2-4x+4)((x^2-6x+9)+1)

\Rightarrow y=(x^2-4x+4)(x^2-6x+10)

\Rightarrow y=x^2x^2-6x^2x+10x^2-4x^2x+4\cdot \:6xx-4\cdot \:10x+4x^2-4\cdot \:6x+4\cdot \:10

\Rightarrow y=x^4-10x^3+14x^2+24x^2-40x-24x+40

\Rightarrow y=x^4-10x^3+38x^2-64x+40

Therefore, this is the required polynomial function.


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