7. Write a paragraph proof of theorem 3-8: in a plane, if two lines are perpendicular to the same line, then they are parallel t
o each other.
1 answer:
Step-by-step explanation:
Assume that two lines x and y are perpendicular to the line z.
+) x intersects z at point A => Angle A1 = 90°
+) y intersects z at point B => Angle B1 = 90°
As it can be seen from the figure that Angle A1 and Angle B1 are corresponding angles and they are congruent.
According to the corresponding angle theorem, when two parallel lines are cut by a transversal, the pair of corresponding angles are congruent and vice versa, when two lines are cut by the same line and the pair of corresponding angles are congruent, they are parallel.
=> x and y are parallel.
You might be interested in
At the start, the tank contains
(0.02 g/L) * (1000 L) = 20 g
of chlorine. Let <em>c</em> (<em>t</em> ) denote the amount of chlorine (in grams) in the tank at time <em>t </em>.
Pure water is pumped into the tank, so no chlorine is flowing into it, but is flowing out at a rate of
(<em>c</em> (<em>t</em> )/(1000 + (10 - 25)<em>t</em> ) g/L) * (25 L/s) = 5<em>c</em> (<em>t</em> ) /(200 - 3<em>t</em> ) g/s
In case it's unclear why this is the case:
The amount of liquid in the tank at the start is 1000 L. If water is pumped in at a rate of 10 L/s, then after <em>t</em> s there will be (1000 + 10<em>t</em> ) L of liquid in the tank. But we're also removing 25 L from the tank per second, so there is a net "gain" of 10 - 25 = -15 L of liquid each second. So the volume of liquid in the tank at time <em>t</em> is (1000 - 15<em>t </em>) L. Then the concentration of chlorine per unit volume is <em>c</em> (<em>t</em> ) divided by this volume.
So the amount of chlorine in the tank changes according to
which is a linear equation. Move the non-derivative term to the left, then multiply both sides by the integrating factor 1/(200 - 5<em>t</em> )^(5/3), then integrate both sides to solve for <em>c</em> (<em>t</em> ):
There are 20 g of chlorine at the start, so <em>c</em> (0) = 20. Use this to solve for <em>C</em> :