Answer:Where is the graph? I'd love to help, but I'd need an image of the graph, please and thank you
Your equation is:
(x - 6)² + (y + 8)² = 100
When you plug in (0, 0) into this equation, you get:
(0 - 6)² + (0 + 8)² = 100
(- 6)² + 8² = 100
36 + 64 = 100
100 = 100
Looking at the numbers, we can see that the differences are 72, 36, and 18. Notice that they are each factors of the next by the same constant 2.
18 * 2 = 36
36 * 2 = 72
There, it means that 18/2 will be subtracted from 20.
20 - 18/2
= 20 - 9
= 11
The next number in the sequence would be 11. Hope this helps!
Answer:
0.6 + 15b + 4 = 25.6 all equivelant
0.6 + 15b + 4 = 25.6 all equivelant
0.6 + 15b + 4 = 25.6 all equivelant
0.6 + 15b + 4 = 25.6 all equivelant
0.6 + 15b + 4 = 25.6 all equivelant
0.6 + 15b + 4 = 25.6 all equivelant
Step-by-step explanation:
0.6 + 15b + 4 = 25.6 all equivelant
0.6 + 15b + 4 = 25.6 all equivelant
0.6 + 15b + 4 = 25.6 all equivelant
0.6 + 15b + 4 = 25.6 all equivelant
0.6 + 15b + 4 = 25.6 all equivelant
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.