ΔJKL, sin(JLK)=JK/KL=3/6=0.5
<JKL=30deg.
cos(JLK)=JL/KL
JL=KL*cos(JLK)=6cos(30deg.)
<JLK + <JLM = <MLK
30 + <JLM = 61
<JLM=31deg
in ΔMJL
tan(JLM)=JM/JL
JM=JL*tan(JLM)
=6cos(30deg.)*tan(31deg)
=3.12
We have that the stream of water is coming out in small quantities; at most liters at a time since the stream has a width of around 10cm at most times. Hence, cubic meters per second is too large a unit to measure the small quantity of water going through the shower.
We have that we can calculate the rate if we know the surface area of the flow and the speed of the water. If one multiplies those 2 together, one gets the rate because the speed of water is pretty much how much a front of water is moving per second; if you multiply it by its surface area, you get how much a volume of water is moving.
Answer: A growing division of labor between employees with different skills
Step-by-step explanation:
Answer:
a random sample of size 5 from a population that is approximately normal
a random sample of size 60 from a population that is strongly skewed to the left.
Step-by-step explanation:
They are both correct
I am the goddes of the air pods and the books of the laptops
