T(1) = 3, t(n) = -2t(n-1) + 1
So t(2) = -2(t(1)) + 1 = -2(3) + 1 = -5
t(3) = -2(t(2)) +1 = -2(-5) +1 = 11
t(4) = -2(t(3)) +1 = -2(11) +1 = -21
t(5) = -2(t(4)) +1 = -2(-21) +1 = 43
Answer:
Step-by-step explanation:
X^2+7x+63=63
<=> x^2+7x=63-63=0
<=> X(x+7)=0
<=> x1=0
X2=-7
24-x/6=-5
I'm not sure if this is correct
Let x = volume of saline solution A,
y = volume of saline solution B.
We're going to set up two equations.
Firstly, x + y = 6. That gives us y = 6 – x (1)
Second, 0.1x + 0.05y = 0.08•6,
or 0.1x + 0.05y = 0.48 (2)
Substitute (1) into (2):
0.1x + 0.05(6 – x) = 0.48
0.1x + 0.3 – 0.05x = 0.48
0.05x + 0.3 = 0.48
Subtract 0.3 from both sides:
0.05x = 0.18
x = 3.6
Then y = 6 – 3.6 = 2.4
So we have to use 3.6 litres of solution A and 2.4 litres of solution B.
