1.

lord [1]

Answer: (9, -2)

Step-by-step explanation:

Let's start with what makes a function even. A function is even if the graph of it is symmetric about the y-axis. What is really means is that if you were to fold your graph paper where the crease is on the y-axis, the graph should be the same on each side.

Now since g is an even function, it's correct for us to assume that it is symmetric about the y-axis. What this means is that we expect to find a value at the same point, except on the right side (because our coordinate is negative).

Our coordinate is (-9, -2). If we were to plot it, we'd see it would be in the 3rd quadrant, or the bottom left one. To be symmetrical on the right side, we know there is a point with the same coords in the 4th quadrant. To be on the right, our x coordinate would be positive, and our y coordinate will stay the same.

5 0

2 years ago

lisa drank 1/3 quart of orange juice and lane drank 1/2 quart of orange juice. how many quarts did they drink together

mezya [45]

Answer:

$\frac{5}{6}$ quarts

Step-by-step explanation:

Total amount of drinks = $\frac{1}{3}$ + $\frac{1}{2}$

You would want to make the denominators the same before adding them, so I will multiply the first fraction by 2 and the second one by 3 so their denominators are 6.

$\frac{1}{3}$ + $\frac{1}{2}$ = $\frac{2}{6}$ + $\frac{3}{6}$

Now we can directly add them together while keeping the denominator the same.

$\frac{2}{6}$ + $\frac{3}{6}$ = $\frac{5}{6}$ quarts

4 0

2 years ago

What is the answer to this slope A (1,3),B (4,7)

jeyben [28]

The slope of these points is 4/3

7 0

3 years ago

Viola says that she can find the surface area of a

Viefleur [7K]

Answer:

This is only true with a cube

Step-by-step explanation:

Only a cube has 6 equal faces of equal size.

I can't really put a diagram here so i hope that explanation helped!

Stay safe! <3

8 0

2 years ago

The slope of the line tangent to the curve y^2 + (xy+1)^3 = 0 at (2, -1) is ...?

Lina20 [59]

$\frac{d}{dx}(y^2 + (xy+1)^3 = 0)
\\
\\\frac{d}{dx}y^2 + \frac{d}{dx}(xy+1)^3 = \frac{d}{dx}0
\\
\\\frac{dy}{dx}2y + \frac{d}{dx}(xy+1)\ *3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{d}{dx}(xy)+\frac{d}{dx}1\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{d}{dx}xy+x\frac{d}{dx}y+\frac{dx}{dx}\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{dx}{dx}y+x\frac{dy}{dx}+\frac{dx}{dx}0\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0$
$\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(3y+3x\frac{dy}{dx}\right) (xy+1)^2 = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) (xy+1)^2 = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) ((xy)^2+2xy+1) = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) ((xy)^2+2xy+1) = 0
\\
\\2y\ y' + 3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y') = 0
\\
\\3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y'+2yy' ) = 0
\\
\\y'(3x +6x^2y+3x^2y+2y ) = -3x^2y^3 -6xy^2-3y

$
$y' = \frac{-3x^2y^3 -6xy^2-3y}{(3x +6x^2y+3x^2y+2y )};\ x=2,y=-1
\\
\\y' = \frac{-3(2)^2(-1)^3 -6(2)(-1)^2-3(-1)}{(3(2) +6(2)^2(-1)+3(2)^2(-1)+2(-1) )}
\\
\\y' = \frac{-3(4)(-1) -6(2)(1)+3}{(6 +6(4)(-1)+3(4)(-1)-2 )}
\\
\\y' = \frac{12 -12+3}{(6 -24-12-2 )}
\\
\\y' = \frac{3}{( -32 )}$

7 0

2 years ago