Answer:
slope of perpendicular line - m = -1/2
passing through (-2,3)
thus,
y-3 = -1/2 (x+2)
or y = -1/2x +2
1. sqrt 52 = 7.211 rounds to 7
2. irrational number
3. sqrt 441 = 21....rational
4. 12^2 + 3^2 = h^2
144 + 9 = h^2
153 = h^2
sqrt 153 = h
12.4 = h <===
5. 6^2 + b^2 = 18^2
36 + b^2 = 324
b^2 = 324 - 36
b^2 = 288
b = sqrt 288
b = 16.97 rounds to 17 <==
6. 39^2 + 52^2 = c^2
1521 + 2704 = c^2
4225 = c^2
sqrt 4225 = c
65 = c
(39 + 52) - 65 = 91 - 65 = 26 miles shorter <==
7. 4^2 + b^2 = 16^2
16 + b^2 = 256
b^2 = 256 - 16
b^2 = 240
b = sqrt 240
b = 15.49 rounds to 15.5 <==
8. ?
Answer:
The decay rate is 5%.
Step-by-step explanation:
Let a substance is decaying at the rate of r% per hour from the initial value of P for t hours, then the final value of the substance is given by the function
........... (1)
Comparing this equation with the original equation given as
............ (2) we get,
⇒
⇒ r = 5%.
Therefore, the decay rate is 5%. (Answer)
Theirs a trick to cheat on khan academy
1) Turn off Wifi
2) Go to hint and get the answer
3) Take picture of answer or remember it
4) Go to inspect element ——> arrows that look like this “>>”—> Application—-> Local Storage (click it until it says khan academy .com etc then click that then go to “X” by value and next to it click that then turn on wifi and put the answer
Unsure of what you are asking!
But if the issue here is how to define a line segment, write what you do know and then reconsider "undefined terms."
A line segment is a straight line that connects a given starting point and given ending point.
If you consider a circle of radius 3 units, the radius can be thought of as the line segment connecting the center of the circle to any point on the circumference of the circle.
If the center of a given circle is at C(0,0) and a point on the circumference is given by R(3sqrt(2),3sqrt(2)), then AC is the line segment joining these two points. This line segment has length 3 and is in the first quadrant, with coordinates x=3sqrt(2) and y=3sqrt(2) describing the end point of the segment.