Answer: Un alambre de acero de 1.5 m se estira 2.0 mm con fuerza F. El diámetro del alambre de acero es 4.0 mm. El módulo de Young del acero es 2.0 x 1011 Nm-2. Determine la fuerza F aplicada sobre el alambre.
Step-by-step explanation:
Answer:
Step-by-step explanation:
I have 30 coins, all nickels, dimes, and quarters, worth $4.60. There are two more dimes than quarters. How many of each kind of coin do I have.
..
let quarters be x
dimes = x+2
...
dimes + quarters = x+x+2=2x+2
...
nickels = 30-(2x+2)
...
5(30-(2x+2))+10(x+2)+25x=460
5(30-2x-2)+10x+20+25x=460
150-10x-10+10x+20+25x=460
160+25x=460
-160
25x=460-160
25x=300
/25
x=300/25
x=12 ---- quarters
x+2= 12+2=14 dimes
30-(2x+2)=4 nickels
...
check
4*5+14*10+12*25=20+140+300=460
X/(x+5) = (x-2)/(x+1)
x(x+1) = (x -2) (x + 5)
x^2 + x = x^2 - 2x + 5x - 10
x = 3x - 10
3x - x = 10
2x = 10
x = 5
answer is A
5
Answer: B and E
Step-by-step explanation:
I just took the assignment. :)
Answer:

Step-by-step explanation:
Given

See attachment
Required
Determine the measure of 
and
are on a straight line.
So:
--- angle on a straight line
Substitute known values

Collect like terms

