A closed, or shaded, circle is used to represent the inequalities greater than or equal to (≥) or less than or equal to (≤) . The point is part of the solution. An open circle is used for greater than (>) or less than (<). The point is not part of the solution.

3 0

3 years ago

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For the equation given below, evaluate y′ at the point (−2,2)<br> xe^y−4y=2x−4−2e^2

jeka94

Implicit differentiation
chain rule is important here

I'll show the steps partially
$e^y+xe^yy'-4y'=2$
$xe^yy'-4y'=2-e^y$
$y'(xe^y-4)=2-e^y$
$y'=\dfrac{2-e^y}{xe^y-4}$
now evaluate for (-2,2)
x=-2 and y=2
$y'=\dfrac{2-e^2}{-2e^2-4}$
$y'=\dfrac{2-e^2}{-2e^2-4}$
$y'=\dfrac{e^2-2}{2e^2+4}$
that's it, simplest form

3 0

4 years ago

PLEASE HELP ASAP WILL GIVE BRAINLIEST!!!!

max2010maxim [7]

Answer:

Step-by-step explanation:

y intercept is -6

the slope is already in it's fraction form but if you want it back to whole number it's y=1.33x-6

the slope is positive

Some plots you can put are (6, 2) and (12, 10)

Just start at (0, -6) and counts 4 up and 3 right

4 0

3 years ago

X^2 - 16 = 0 by extracting the square root

scoray [572]

Answer:

{-4, 4}

Step-by-step explanation:

x² - 16 = 0
x² = 16
x = √16
x = ± 4

6 0

2 years ago