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3 years ago

24xy-xy^3+x^2 find the degree plz

Mathematics

1 answer:

Rzqust [24]3 years ago

3 0

Answer:

Step-by-step explanation:

24xy---> 2

-xy3--->4

x2--->2

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4 years ago

X²+y²-6x+14y-1=0<br> and please show your work so I can learn

Eva8 [605]

Hello there,

I hope you and your family are staying safe and healthy during this winter season.

$x^2 + y^2 -6x+14y-1=0$

We need to use the Quadratic Formula*

$x =\frac{-b+\sqrt{b^2}-4ac }{2a}$ , $\frac{-b-\sqrt{b^2} -4ac }{2a}$

Thus, given the problem:

$a = 1, b=-6, c=y^2+14y-1$

So now we just need to plug them in the Quadratic Formula*

$x=\frac{6+2\sqrt{(-6)^2-4(y^2+14y-1)} }{2}$ , $x=\frac{6-\sqrt{(-6)^2-4(y^2+14y-1)} }{2}$

As you can see, it is a mess right now. Therefore, we need to simplify it

$x=\frac{6+2\sqrt{10-y^2-14y} }{2}$ , $x = \frac{6-2\sqrt{10-y^2-14y} }{2}$

Now that's get us to the final solution:

$x=3+\sqrt{10-y^2-14y}$ , $x=3-\sqrt{10-y^2-14y}$

It is my pleasure to help students like you! If you have additional questions, please let me know.

Take care!

~Garebear

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3 years ago

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