hectictar

\((\frac19)^{\frac32}=\sqrt{\frac19^3}=\sqrt{\frac19\cdot\frac19\cdot\frac19}=\frac19\sqrt{\frac19}=\frac19\cdot\frac{\sqrt1}{\sqrt9}=\frac{1}{9}\cdot\frac13\mathbf{=\frac{1}{27}}\)

Thank you! I appreciate it 

This is a further edit of RS's version.

Purple is my writing.

Here are the landforms: condensation, precipitation, collection, and evaporation. The water cycle is the cycle of processes by which water circulates between the earth's oceans, atmosphere, and land, and involves precipitation as rain and snow, drainage in streams and rivers, and a return to the atmosphere by evaporation and transpiration. Dew is not rain; it is the condensation on grass. About ninety-five percent of the earth's surface is water. The remaining five percent is land. The amount of water never changes on Earth. The water that we drink today is animal or human saliva that has been purified and made ready to use again. Earth's water never came from another planet, it was just there when Earth was created. Water that has turned into water vapor spends about ten days floating in the air.

I tried to change it only very little. :)

Here is a top down view of the table.

The purple semicirles are the leaves. The blue square is the table.

As you can see, the table can be split up into two shapes: a square and a circle.

The area without the drop leaves = the area of the square.

The area with the drop leaves = the area of the square + the area of the circle.

So... let's start by finding the area of each shape.

the area of the square = length * width = 4 * 4 = 16 sq. ft.

the area of the circle = π * radius² = π * (4/2)² = π * 2² = 4π sq. ft.

So..

The area without the drop leaves = the area of the square = 16 sq. ft.

The area with the drop leaves = the area of the square + the area of the circle. = 16 + 4π ≈ 28.566 sq. ft.

You can make it either one by selecting one of the bubbles in the bottom left corner. It should be in degrees by default. :)

\(\frac{d-1}{5}=\frac{d-3}{3}\)

Cross-multiply.

\((3)(d-1)=(5)(d-3)\)

Distribute.

\(3d-3=5d-15\)

Subtract 3d from both sides.

\(-3=5d-3d-15\)

Add 15 to both sides.

\(-3+15=5d-3d \\~\\ 12=2d\)

Divide both sides by 2.

\(\frac{12}{2}=d \\~\\ 6=d\)

To solve for x:

\(\tan 60^{\circ}=\frac{35}{x} \\~\\ \frac{\sin60^{\circ}}{\cos60^{\circ}}=\frac{35}{x}\)

You can look at a unit circle to get exact values for sin60º and cos60º .

\(\frac{\frac{\sqrt3}{2}}{\frac12}=\frac{35}{x} \\~\\ \frac{\sqrt3}{2}*2=\frac{35}{x} \\~\\ \sqrt3=\frac{35}{x} \\~\\ x=\frac{35}{\sqrt3} \\~\\ x=\frac{35\sqrt3}{3}\)

Factor the numerator and denominator.

\(\frac{x^2-2x-8}{x^2-16}=\frac{(x-4)(x+2)}{(x-4)(x+4)}\)

Now it is easy to see that if x = -4 or +4, the denominator will = 0.

(If it wasn't easy to see, just set the denominator equal to zero and solve for x.)

So there is a restriction that x  ≠ ± 4

Now reduce the numerator and denominator by (x - 4) .

\(\frac{(x-4)(x+2)}{(x-4)(x+4)}=\frac{(x+2)}{(x+4)} \\~\\ \frac{x+2}{x+4}\)

Now it is simplified, and the only restriction is x  ≠ ± 4   

Don't worry..I'll keep a watch out for any suspicious activity...

I got you bro