Its A
3/7 + 2/3 = 23/21
and 23/21 as a mixed number is:
1 and 2/21
You should have drawn1 - x-axis and y-axis in light pencil.2 - graphed a down-facing parabola with the top of the frown on the y-axis at y = 2. It should be crossing the x-axis at ±√2. This should be in dark pencil or another color.3 - In dark pencil or a completely new color, draw a rectangle with one of the horizontal sides sitting on top of the x-axis and the other horizontal side touching the parabola at each of the top corners of the rectangle. The rectangle will have half of its base in the positive x-axis and the other half on the negative x-axis. It should be split right down the middle by the y-axis. So each half of the base we will say is "x" units long. So the whole base is 2x units long (the x units to the right of the y-axis, and the x units to the left of the y-axis) I so wish I could draw you this picture... In the vertical direction, both vertical edges are the same length and we will call that y. The area that we want to maximize has a width 2x long, and a height of y tall. So A = 2xy This is the equation we want to maximize (take derivative and set it = 0), we call it the "primary equation", but we need it in one variable. This is where the "secondary equation" comes in. We need to find a way to change the area formula to all x's or all y's. Since it is constrained to having its height limited by the parabola, we could use the fact that y=2 - x2 to make the area formula in only x's. Substitute in place of the "y", "2 - x2" into the area formula. A = 2xy = 2x(2 - x2) then simplify A = 4x - 2x3 NOW you are ready to take the deriv and set it = 0 dA/dx = 4 - 6x2 0 = 4 - 6x2 6x2 = 4 x2 = 4/6 or 2/3 So x = ±√(2/3) Width remember was 2x. So the width is 2[√(2/3)]Height is y which is 2 - x2 = 2 - 2/3 =4/3
Given:
The expression is
![\sqrt[3]{48}=\sqrt[3]{8\cdot \_\_}=](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B48%7D%3D%5Csqrt%5B3%5D%7B8%5Ccdot%20%5C_%5C_%7D%3D)
To find:
The simplified form of the expression.
Solution:
We have,
The expression ![\sqrt[3]{48}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B48%7D)
can be written as
![\sqrt[3]{48}=\sqrt[3]{8\cdot 6}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B48%7D%3D%5Csqrt%5B3%5D%7B8%5Ccdot%206%7D)
![\sqrt[3]{48}=\sqrt[3]{8}\cdot \sqrt[3]{6}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B48%7D%3D%5Csqrt%5B3%5D%7B8%7D%5Ccdot%20%5Csqrt%5B3%5D%7B6%7D)
![[\because \sqrt[3]{ab}=\sqrt[3]{a}\sqrt[3]{b}]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%5Csqrt%5B3%5D%7Bab%7D%3D%5Csqrt%5B3%5D%7Ba%7D%5Csqrt%5B3%5D%7Bb%7D%5D)
![\sqrt[3]{48}=2\cdot \sqrt[3]{6}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B48%7D%3D2%5Ccdot%20%5Csqrt%5B3%5D%7B6%7D)
![\sqrt[3]{48}=2\sqrt[3]{6}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B48%7D%3D2%5Csqrt%5B3%5D%7B6%7D)
Therefore, ![\sqrt[3]{48}=\sqrt[3]{8\cdot 6}=2\sqrt[3]{6}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B48%7D%3D%5Csqrt%5B3%5D%7B8%5Ccdot%206%7D%3D2%5Csqrt%5B3%5D%7B6%7D)
.
C. 100 because a quadrilateral must add up to 360. :)
I believe the slope is -1
