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Rus_ich [418]
3 years ago
12

Pls i need it like now

Mathematics
1 answer:
schepotkina [342]3 years ago
7 0
You need to put the x or y value in the other equation ;
For example excersize a
It given that
Y=x^2+3x-1
Then you put it on the first equation :
x+x^2+3x-1=4
And then you solve it
x+x^2+3x-1-4=0
4x+x^2-5=0
We will make the order to seem easier
x^2+4x-5=0
x1=-5
x2=1
Then you put the x1,2 that you found on the second equation :
y=(-5)^2+3*(-5)-1=9
y=1^2+3*1-1=3
For summary :
x1=-5
x2=1
y1=9
y2=3

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The length of the space diagonal can be found to be the root of the squares of the three orthogonal edge lengths. For a cube, those edge lengths are all the same, so the diagonal length is ...

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when the prism is a cube, these are all the same (a=b=c). This is the formula we used above.

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Simplify<br> 7^2x^-3y/49x^-3y^-2
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\frac{7^2x^{-3}}{49x^{-3} y^{-2}} \\ \\  \frac{49x^{-3}y}{49x^{-3}y^{-2}} \\ \\  \frac{49x^{-3}y}{49x^{-3} \times  \frac{1}{y^2} } \\ \\  \frac{49x^{-3}y}{ \frac{49x^{-3}}{y^2} } \\ \\ 49x^{-3} y \times  \frac{y^2}{49x^{-3}} \\ \\  \frac{x^{-3}}{x^{-3}} yy^2 \\ \\ 1 \times yy^2 \\ \\ yy^2 \\ \\ y^3 \\ \\

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