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IrinaK [193]
3 years ago
15

Select the correct domain and range of the relation. (-4, -4)(-1,3)(0,5)(3,9)

Mathematics
1 answer:
DaniilM [7]3 years ago
6 0

Answer:

Domain: {1, 2, 3, 4}

Range: {3, 6, 9, 12}

Step-by-step explanation:

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ax^2+bx+c=0\\\\\text{From quadratic formula the roots are equal:}\\\\x_1=\dfrac{-b-\sqrt{b^2-4ac}}{2a}\ \text{and}\ x_2=\dfrac{-b+\sqrt{b^2-4ac}}{2a}\\\\\text{We have}\ 2x^2+7x-15=0\\\\a=2,\ b=7,\ c=-15\\\\\text{and}\ r,\ s\ \text{are two solution}\\\\\sqrt{b^2-4ac}=\sqrt{7^2-4(2)(-15)}=\sqrt{49+120}=\sqrt{169}=13\\\\x_1=\dfrac{-7-13}{2(2)}=\dfrac{-20}{4}=-5\\\\x_2=\dfrac{-7+13}{2(2)}=\dfrac{6}{4}=\dfrac{3}{2}\\\\\dfrac{3}{2}>-5\ \text{therefore}\ r=\dfrac{3}{2}\ \text{and}\ s=-5.

r-s=\dfrac{3}{2}-(-5)}=\dfrac{3}{2}+5=\dfrac{3}{2}+\dfrac{10}{2}=\dfrac{13}{2}\\\\Answer:\ \boxed{B)\ \dfrac{13}{2}}

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