Problem 1
Draw a straight line and plot P anywhere on it. Use the compass to trace out a faint circle of radius 8 cm with center P. This circle crosses the previous line at point Q.
Repeat these steps to set up another circle centered at Q and keep the radius the same. The two circles cross at two locations. Let's mark one of those locations point X. From here, we could connect points X, P, Q to form an equilateral triangle. However, we only want the 60 degree angle from it.
With P as the center, draw another circle with radius 7.5 cm. This circle will cross the ray PX at location R.
Refer to the diagram below.
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Problem 2
I'm not sure why your teacher wants you to use a compass and straightedge to construct an 80 degree angle. Such a task is not possible. The proof is lengthy but look up the term "constructible angles" and you'll find that only angles of the form 3n are possible to make with compass/straight edge.
In other words, you can only do multiples of 3. Unfortunately 80 is not a multiple of 3. I used GeoGebra to create the image below, as well as problem 1.
You can show that 
by constructing a triangle.
Take two points, O(0, 0) and A(1, 0), and let B be the point on the unit circle such that the angle between the line segments OA and OB is 
radians.
Since both A and B lie on the circle, the line segments OA and OB both have length 1 (same as the circle's radius). We finish constructing the triangle by connect A and B.
Since OB and OA have the same length, triangle OAB is isosceles, but more than that, it's also equilateral. Why? Because the interior angles of any triangle always add to 
radians. We know one of the angles is radians, which leaves a contribution of 
radians between the remaining angles A and B. Angles A and B must be congruent (because OAB is isosceles), which means they also have measure radians.
Next, draw an altitude of the triangle through point B, and label the point where it meets the "base" OA, C. Since OAB is equilateral, the altitude BC is also a perpendicular bisector. That means OC has length 
, and by definition of 
we have
Answer:
C(t)=5000 -10t
Step-by-step explanation:
There are many examples in the real world of relationships that are functions.
For example, imagine a tank full of water with a capacity of 5000 liters, this tank has a small hole, by which 10 liters of water are lost every hour.
If we call C the amount of water in the tank as a function of time, then we can write the following equation for C:
Where:
C (t): Amount of water in the tank as a function of time
: Initial amount of water in the tank at time t = 0
a: amount of water lost per hour
t: time in hours
Then the equation is:
The graph of C (t) is a line of negative slope. This relation is a function since for each value of t there is a single value of C.
Its domain is the set of all positive real numbers t between [0,500]
Because the time count starts at t = 0 when the tank is full and ends at t = 500 when empty
Its Range is the set of all positive real numbers C between [0,5000] Because the amount of water in the tank can never be less than zero or greater than 5000Litres
Answer:
-1 = (-3/5)-15 + b
(I don't know the y-intercept bc I don't have a graphing calc but to find it out you just graph it and then see where the line passes thru the y axis)
Answer:
6/15 divided by 6
Step-by-step explanation:
