Question

(x^2y)^4 please answer this question faster​

Answer 1

Answer:

x^8y.........................

Answer 2

Given:

x

+

2

y

=

4

Note that this equation is linear, since all of the terms in

x

and

y

are of degree at most

1

.

We can find the line's

x

intercept by setting

y

=

0

, or equivalently covering up the

y

term to find:

x

=

4

So the intersection with the

x

axis (which has equation

y

=

0

) is the point

(

4

,

0

)

.

Similarly, we can find the line's

y

intercept by setting

x

=

0

, or equivalently covering up the

x

term to find:

2

y

=

4

and hence:

y

=

2

So the intersection with the

y

axis (which has equation

x

=

0

) is the point

(

0

,

2

)

.

Now we can draw our line through those two points:

graph{(x+2y-4)((x-4)^2+y^2-0.02)(x^2+(y-2)^2-0.02) = 0 [-7.58, 12.42, -4.04, 5.96]}