coordinate of B= (2+3,0) = (5,0)
by pythagoras' theorem,
height of triangle^2 + half length of AB^2 = AC^2
height of triangle^2 = 3^2-1.5^2
height of triangle ^2= 6.75
height of triangle= √6.75
coordinates of C= (1.5,√6.75)
Answer:
v = 1/(1+i)
PV(T) = x(v + v^2 + ... + v^n) = x(1 - v^n)/i = 493
PV(G) = 3x[v + v^2 + ... + v^(2n)] = 3x[1 - v^(2n)]/i = 2748
PV(G)/PV(T) = 2748/493
{3x[1 - v^(2n)]/i}/{x(1 - v^n)/i} = 2748/493
3[1-v^(2n)]/(1-v^n) = 2748/493
Since v^(2n) = (v^n)^2 then 1 - v^(2n) = (1 - v^n)(1 + v^n)
3(1 + v^n) = 2748/493
1 + v^n = 2748/1479
v^n = 1269/1479 ~ 0.858
Step-by-step explanation:
Answer:
x = 8, y = 2
Step-by-step explanation:
Multiply the second equation by -2:
x + 6y = 20
-2x - 6y = -28
Add the equations and simplify:
-x = -8
x = 8
Plug x = 8 back into the first equation and solve for y:
8 + 6y = 20
6y = 12
y = 2
They all 3 have no solution!