Which equation can be used to solve for the rate, r, for Sam to change all the light bulbs in the office building by himself?
2 answers:
Answer:
Step-by-step explanation:
Given question is incomplete; here is the complete question.
Working together, Sam and Charlie can change all light bulbs in the office building in 3 hours. It would have taken Charlie 8 hours to do the job alone. Which equation can be be used to solve for the rate 'r', for Sam to change all the light bulbs in the office building by himself.
Let Sam can change the bulbs in the office alone in 'r' hours
Per hour work done by Sam alone =
Let Sam takes he time to change the same bulbs in the office alone = 8 hours
Per hour work done by Sam =
Since, Sam and Charlie complete the work together in time = 3 hours
Therefore, per hour work done by both =
Equation for per work done by both will be,
Per hour work done by both = Per hour work done by Sam alone + Per hour work done by Charlie
Therefore, equation for the rate 'r' will be,
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Problem 1)
AC is only perpendicular to EF if angle ADE is 90 degrees
(angle ADE) + (angle DAE) + (angle AED) = 180
(angle ADE) + (44) + (48) = 180
(angle ADE) + 92 = 180
(angle ADE) + 92 - 92 = 180 - 92
angle ADE = 88
Since angle ADE is actually 88 degrees, we do NOT have a right angle so we do NOT have a right triangle
Triangle AED is acute (all 3 angles are less than 90 degrees)
So because angle ADE is NOT 90 degrees, this means AC is NOT perpendicular to EF
-------------------------------------------------------------
Problem 2)
a) The center is (2,-3)
The center is (h,k) and we can see that h = 2 and k = -3. It might help to write (x-2)^2+(y+3)^2 = 9 into (x-2)^2+(y-(-3))^2 = 3^3 then compare it to (x-h)^2 + (y-k)^2 = r^2
---------------------
b) The radius is 3 and the diameter is 6
From part a), we have (x-2)^2+(y-(-3))^2 = 3^3 matching (x-h)^2 + (y-k)^2 = r^2
where
h = 2
k = -3
r = 3
so, radius = r = 3
diameter = d = 2*r = 2*3 = 6
---------------------
c) The graph is shown in the image attachment. It is a circle with center point C = (2,-3) and radius r = 3.
Some points on the circle are
A = (2, 0)
B = (5, -3)
D = (2, -6)
E = (-1, -3)
Note how the distance from the center C to some point on the circle, say point B, is 3 units. In other words segment BC = 3.
AC is only perpendicular to EF if angle ADE is 90 degrees
(angle ADE) + (angle DAE) + (angle AED) = 180
(angle ADE) + (44) + (48) = 180
(angle ADE) + 92 = 180
(angle ADE) + 92 - 92 = 180 - 92
angle ADE = 88
Since angle ADE is actually 88 degrees, we do NOT have a right angle so we do NOT have a right triangle
Triangle AED is acute (all 3 angles are less than 90 degrees)
So because angle ADE is NOT 90 degrees, this means AC is NOT perpendicular to EF
-------------------------------------------------------------
Problem 2)
a) The center is (2,-3)
The center is (h,k) and we can see that h = 2 and k = -3. It might help to write (x-2)^2+(y+3)^2 = 9 into (x-2)^2+(y-(-3))^2 = 3^3 then compare it to (x-h)^2 + (y-k)^2 = r^2
---------------------
b) The radius is 3 and the diameter is 6
From part a), we have (x-2)^2+(y-(-3))^2 = 3^3 matching (x-h)^2 + (y-k)^2 = r^2
where
h = 2
k = -3
r = 3
so, radius = r = 3
diameter = d = 2*r = 2*3 = 6
---------------------
c) The graph is shown in the image attachment. It is a circle with center point C = (2,-3) and radius r = 3.
Some points on the circle are
A = (2, 0)
B = (5, -3)
D = (2, -6)
E = (-1, -3)
Note how the distance from the center C to some point on the circle, say point B, is 3 units. In other words segment BC = 3.