Answer:
Original position: base is 1.5 meters away from the wall and the vertical distance from the top end to the ground let it be y and length of the ladder be L.
Step-by-step explanation:
By pythagorean theorem, L^2=y^2+(1.5)^2=y^2+2.25 Eq1.
Final position: base is 2 meters away, and the vertical distance from top end to the ground is y - 0.25 because it falls down the wall 0.25 meters and length of the ladder is also L.
By pythagorean theorem, L^2=(y -0.25)^2+(2)^2=y^2–0.5y+ 0.0625+4=y^2–0.5y+4.0625 Eq 2.
Equating both Eq 1 and Eq 2: y^2+2.25=y^2–0.5y+4.0625
y^2-y^2+0.5y+2.25–4.0625=0
0.5y- 1.8125=0
0.5y=1.8125
y=1.8125/0.5= 3.625
Using Eq 1: L^2=(3.625)^2+2.25=15.390625, L=(15.390625)^1/2= 3.92 meters length of ladder
Using Eq 2: L^2=(3.625)^2–0.5(3.625)+4.0625
L^2=13.140625–0.90625+4.0615=15.390625
L= (15.390625)^1/2= 3.92 meters length of ladder
<em>hope it helps...</em>
<em>correct me if I'm wrong...</em>
<span>FG = EG−EF
= 10 − 4
= 6</span>
Hello from MrBillDoesMath!
Answer:
x = 1/2 (1 +\- i sqrt(23))
Discussion:
x \3x - 2 = (x/3)*x - 2 = (x^2)/3 - 2 (*)
1 \3x - 4 = (1/3)x - 4 (**)
(*) = (**) =>
(x^2)/3 -2 = (1/3)x - 4 => multiply both sides by 3
x^2 - 6 = x - 12 => subtract x from both sides
x^2 -x -6 = -12 => add 12 to both sides
x^2-x +6 = 0
Using the quadratic formula gives:
x = 1/2 (1 +\- i sqrt(23))
Thank you,
MrB
