B.Number of eye blinks in a minute.
We do not have a set amount of blinks per minute.
A.Inccorect,a touchdown is 6 points
C.Speed of a race car rises at a constant rate depending on the car,speed e.t.c
D.Dogs get placed in a kennel and have a set ammount.
Answer for 11-13: m<2: 150, m<6: 150, m<7: 150
Step-by-step explanation:
There is 180 degrees in a straight line. If one part of the line (angle) is 30 degrees, then the other part is 150. If you look at the image, you would see that m<2 is congruent to m<6 ,which means same, and m<7 is congruent to m<3.
Answer for 14-16: m<EBG: 150, m<AGH: 150, m<DHF: 30.
Step-by-step explanation:
4x = 150 2x + 50 = 150
x = 37.5 2x = 150 - 50
4(37.5) = 150 2x = 100
x = 50
2(50) + 50 =
100 + 50 = 150
Answer:
2.09.ft/min
Step-by-step explanation:
There is a formula for the linear velocity using rpm:
V = (2π)/60 * r * rpm
V = (2π)/60 * 2ft * 10 = 2.09ft/min
UUUUhhhhhh....i cant really see it sorry.
Answer:
Here we can use the relationship:
Distance = time*speed.
When Jose walks, his speed is 4 mph.
Then if he walks for X hours, the distance that he will travel is:
D = 4mph*X
When Jose runs, his speed is 8mph.
Then if he runs for T hours, the distance that he will travel is:
D´ = 8mph*Y
And we know that he travels in total 20 miles, then we must have that:
D + D´= 20mi
This leads to:
4mph*X + 8mph*Y = 20mi
Where X is the time that he walked, and Y is the time that he runed.
Then the equation that represents the different amounts of times that Jose runs and walks is:
4mph*X + 8mph*Y = 20mi
Where we can not really find the solutions for Y and X, because there is only one equation and two variables.
