@pinky try these
Lesson 7 Circle Unit Test.
Here you go:
1- 19.6
2- 78
3-243
4-53.0
5-53.5
6-73
7-31
8-54
9-15.8
10-35.9
11-C (circle on the graph on the left hand side)
12-28
13-11.6
14-56
15-26616-x=15 & triangle is isoscles
17- Paraphrase please!- The center of the circle described by the equation is (-6,-4), so that is the position of the source. The radius of the circle is the square root of 36, which is 6. This means that the range of the radio signal extends out 6 units from the source in all directions. Therefore, any point on the circle with center (-6,-4) and radius 6 is within range of the radio signal.
18- A.((-10+4)/2,(-2+9)/2)
(-3,3.5)
B. sqrt ( ( -10-(-3))^2+(-2-3.5)^2)
sqrt (49+31.25)
sqrt(80.25)
C. (x+3)^2+(y-3.5)^2=80.25
Thank you, I will try these answers!
Here are the step-by-step solutions for the given answers:
1. Value of 1 - 19.6
2. Value of 2 - 78
3. Value of 3 - 243
4. Value of 4 - 53.0
5. Value of 5 - 53.5
6. Value of 6 - 73
7. Value of 7 - 31
8. Value of 8 - 54
9. Value of 9 - 15.8
10. Value of 10 - 35.9
11. Value of 11 - Option C (a circle on the graph on the left-hand side)
12. Value of 12 - 28
13. Value of 13 - 11.6
14. Value of 14 - 56
15. Value of 15 - 266
- Solve for x: x = 15
- The triangle is isosceles.
17. Paraphrase of 17 - The center of the circle described by the equation is (-6,-4), indicating the source's position. The radius of the circle is the square root of 36, equal to 6. Hence, the radio signal range extends 6 units from the source in all directions. Consequently, any point on the circle centered at (-6,-4) with a radius of 6 is within the radio signal range.
18. Value of 18:
A. Calculate the coordinates: ((-10+4)/2, (-2+9)/2) = (-3,3.5)
B. Calculate the distance:
- d = √((-10-(-3))^2 + (-2-3.5)^2)
- d = √(49 + 31.25)
- d = √(80.25)
C. Calculate the equation: (x+3)^2 + (y-3.5)^2 = 80.25
I hope this helps!