Given:
4log1/2^w (2log1/2^u-3log1/2^v)
Req'd:
Single logarithm = ?
Sol'n:
First remove the parenthesis,
4 log 1/2 (w) + 2 log 1/2 (u) - 3 log 1/2 (v)
Simplify each term,
Simplify the 4 log 1/2 (w) by moving the constant 4 inside the logarithm;
Simplify the 2 log 1/2 (u) by moving the constant 2 inside the logarithm;
Simplify the -3 log 1/2 (v) by moving the constant -3 inside the logarithm:
log 1/2 (w^4) + 2 log 1/2 (u) - 3 log 1/2 (v)
log 1/2 (w^4) + log 1/2 (u^2) - log 1/2 (v^3)
We have to use the product property of logarithms which is log of b (x) + log of b (y) = log of b (xy):
Thus,
Log of 1/2 (w^4 u^2) - log of 1/2 (v^3)
then use the quotient property of logarithms which is log of b (x) - log of b (y) = log of b (x/y)
Therefore,
log of 1/2 (w^4 u^2 / v^3)
and for the final step and answer, reorder or rearrange w^4 and u^2:
log of 1/2 (u^2 w^4 / v^3)
Answer:
............................
<span>Draw a regular hexagon. Connect the center to each of the six vertices. Thus, you have six triangles, each with base 10.. The apothem is the height of each triangle. Then the area of each triangle is (1/2)(10)(12) = 60. You have six triangles so the aarea of the hexagon is 6*60 = 360.</span>
Let's call L the width of the rectangle and W its width. The area of the rectangle is the product between the length and the width, and we are also told that the area is 300 square meters, so we can write

Moreover, we know that the length is 5 meters longer than the width:

We have a system of 2 equations in 2 unknown variables, L and W. If we substitute the second equation into the first one, we get


which has two solutions: W=-20 and W=15. We can discard the negative solution since it does not have physical meaning, and now we can substitute the value of W into the second equation to find L:

<span>Therefore, the rectangle has width 15 meters and length 20 meters.</span>
Answer:
Perpendicular lines.
Perpendicular lines are lines that intersect at a right (90 degrees) angle.