Answer:
Part (A) The required volume of the column is 
.
Part (B) The volume be 
.
Step-by-step explanation:
Consider the provided information.
It is given that the we have a square with side length "s" lies in a plane perpendicular to a line L.
Also One vertex of the square lies on L.
Part (A)
Suppose there is a square piece of a paper which is attached with a wire through one corner. As you blow it up it spins around on the wire.
This square moves a distance h along L, and generate a corkscrew-like column with square.
The cross section will remain the same.
So the cross section area of original column and the cross section area of twisted column at each point will be the same.
The volume of the column is the area of square times the height.
This can be written as:
Hence, the required volume of the column is .
Part (B) What will the volume be if the square turns twice instead of once?
If the square turns twice instead of once then the volume will remains the same but divide the volume into two equal part.
Hence, the volume be .
7. X2=25
X=5, x=—5
8. (X—3)2=49
X=10, x=—4
9. X2+3x—28.
X=4, x=—7.
10. 5x2–8=3x.
X=8/5, x=—1
Answer:
p=7Q/2
Step-by-step explanation:
Original number of students:
p students to do 1 job in 25 days.
Let r= the rate for 1 student.
pr*25=1
pr*25=1 is the work rate equation for p students.
Lesser number of students:
p-Q students came to do the job and time required was 35 days.
(P-Q)*r*35=1.
The unknowns are p, Q and r
Equate the original number of students and the lesser number of students
pr*25=(P-Q)*r*35
25rp=35rp - 35Qr
Collect like terms
25rp-35rp = -35Qr
Divide both sides by -5
-5rp+7rp=- 7rp
It can be re written as
7rp-5rp=-7Qr
2rp=7Qr
Make p the subject of the formula
p=7Qr/2r
p=7Q/2
p=7Q/2 is the original number of students
-10rp = -35Qr
The system of these two equations can be solved for p. See the THREE unknown
variables, p, r, and Q. You might assume that either r or Q would be a constant.
A I think I’m sorry if I’m wrong haven’t done this in a while
Assuming that the side of the triangle marked 13.5 is tangent to the circle, the there's a right angle where the radius meets the tangent segment. The triangle is then a right triangle and the Pythagorean Theorem applies.
