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3 years ago

Reduce this algebraic fraction. y^3 -3y^2+y-3 / y^2-9

Mathematics

2 answers:

blondinia [14]3 years ago

6 0

Answer:

y³ - 3y² + 1/y+3

Step-by-step explanation:

y² - 9 = (y - 3) ( y + 3)

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

Luden [163]3 years ago

3 0

Answer:

y^2+1/y^2+3

Step-by-step explanation:

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AlexFokin [52]

Answer: 7. g = -50, 8. f= 340/6 9. 3 10 .125

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3 years ago

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What 2 numbers add to get 2 and multiply to 4

rusak2 [61]

Factor 4
4=1 times 4
2 times 2
they don't add to 2
set up equation
x+y=2
xy=4

first equation, subtract x from both sides
y=2-x
subsitute for y
x(2-x)=4
distribute
2x-x^2=4
add x^2
2x=x^2+4
subtract 2x
0=x^2-2x+4
use quadratic formula which is
if you have ax^2+bx+c=0 then
x= $\frac{ -b+/-\sqrt{b^{2}-4ac} }{2a}$ so

1x^2-2x+4=0
a=1
b=-2
c=4
x= $\frac{ -(-2)+/-\sqrt{(-2)^{2}-4(1)(4)} }{2(1)}$
x= $\frac{ 2+/-\sqrt{4-16} }{2}$
x= $\frac{ 2+/-\sqrt{-12} }{2}$
we have $\sqrt{-12}$ and that doesn't give a real solution
therefor there are no real solutions
but if you want to solve fully
x= $\frac{ 2+/-2\sqrt{-3} }{2}$
i= $\sqrt{-1}$
x= $\frac{ 2+/-2i\sqrt{3} }{2}$
x= $1+/-i\sqrt{3}$
x= $1-i\sqrt{3}$ or x= $1+i\sqrt{3}$ (those are the 2 numbers)

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4 years ago

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nevsk [136]

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vodomira [7]

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