We use the Work formula to solve for the unknown in the problem which is W = F x d. First, we solve for the Net Force acting on the car. The Net Force is the summation of all forces acting on the object. For this case, we assume that Friction Force is negligible thus the Net Force is equal to:
F = mgsinα in terms of SI units and in terms of english units we have F = m(g/g₀)(sin α) where g₀ is the proportionality factor, 32.174 ft lb-m / lb-f s²
F = 2500 (32.174/32.174) (sin 12°) = 519.78 lb
W = Fd = 519.78 lb (400 ft) = 207912 ft - lb or 20800 ft-lb
Answer:
Father's age is 50, Son's age is 10
Step-by-step explanation:
X+Y=60
X-Y= 40
Let X represent the Father's age while Y represents the Son's age;
therefore, from the two equations, the value of X is
X= 60-Y
X= 40+Y then putting in the value of x
(40+Y) + Y= 60,
40+2Y=60
2Y=60-40 =20
Y= 10,
Now that we know the value of Y, we can get the value of x
If X+Y=60, then
X+10=60,
X=60-10,
X= 50
Answer:
Step-by-step explanation:
y=-1/3x+9
This is written in the format y=mx+b, where m is the slope and b the y-intercept (the value of y when x=0).
Perpendicular lines have a slope that is the negative inverse of the slope of the reference line, in this case -(1/3). The new slope is 3. equation is:
y = 3x + b
To find b, enter the given point, (-6,-2) and solve for b:
y = 3x + b
-2 = 3(-6) + b
-2 = -18 + b
b = 16
The full equation is y=3x + 18
Answer:
the EXACT answer is 29.16666667
Step-by-step explanation:
First you add 50, 60, and 65 and get 175
she is driving for 6 hours so you divide 175 by six to get the average
Then you get 29.16666667
Answer:
<h2>2/5</h2>
Step-by-step explanation:
The question is not correctly outlined, here is the correct question
<em>"Suppose that a certain college class contains 35 students. of these, 17 are juniors, 20 are mathematics majors, and 12 are neither. a student is selected at random from the class. (a) what is the probability that the student is both a junior and a mathematics majors?"</em>
Given data
Total students in class= 35 students
Suppose M is the set of juniors and N is the set of mathematics majors. There are 35 students in all, but 12 of them don't belong to either set, so
|M ∪ N|= 35-12= 23
|M∩N|= |M|+N- |MUN|= 17+20-23
=37-23=14
So the probability that a random student is both a junior and social science major is
=P(M∩N)= 14/35
=2/5
