Answer:
the first one
Step-by-step explanation:
A graph showing the Earliest Start Times (EST) for project tasks is computed left to right based on the predecessor task durations. For dependent tasks, the earliest start time will be the latest of the finish times of predecessor tasks.
The first graph appears to appropriately represent the table values, using edges to represent task duration, and bubble numbers to represent start times.
The second graph does not appropriately account for duration of predecessor tasks.
The third graph seems to incorrectly compute task completion times (even if you assume that the edge/bubble number swap is acceptable).
(a) Yes all six trig functions exist for this point in quadrant III. The only time you'll run into problems is when either x = 0 or y = 0, due to division by zero errors. For instance, if x = 0, then tan(t) = sin(t)/cos(t) will have cos(t) = 0, as x = cos(t). you cannot have zero in the denominator. Since neither coordinate is zero, we don't have such problems.
---------------------------------------------------------------------------------------
(b) The following functions are positive in quadrant III:
tangent, cotangent
The following functions are negative in quadrant III
cosine, sine, secant, cosecant
A short explanation is that x = cos(t) and y = sin(t). The x and y coordinates are negative in quadrant III, so both sine and cosine are negative. Their reciprocal functions secant and cosecant are negative here as well. Combining sine and cosine to get tan = sin/cos, we see that the negatives cancel which is why tangent is positive here. Cotangent is also positive for similar reasons.
Answer:
Perimeter=25 units
Area= 43 units^2
Step-by-step explanation:
one side = 5 units so perimeter is 5x5=25
split into 5 equal triangles
base =5, height =3.44, formula for area=(base)(height)(0.5)
area of each triangle =(5)(3.44)(0.5)=8.6
entire area=5(8.6)=43
27 divided by 15 is 1.8, so that should be the correct answer.
Answer:
2000 L
Step-by-step explanation:
There are 1250 L of water in a tank at present. If the tank is 0.625 full, what is the capacity of the tank?
The simple solution is:
1250 L ÷ 0.625 = 2000 L
The algebraic solution is:
Let <em>c</em> equal the capacity of the tank.
Therefore, <em>c</em> × 0.625 = 1250.
Divide both sides by 0.625:
<em>c</em> × 0.625 ÷ 0.625 = 1250 ÷ 0.625
And simplify:
<em>c</em> = 1250 ÷ 0.625
<em>c</em> = 2000
