Answer:
0.15 of a walk
Step-by-step explanation:
2/60 =0.15
Therefore you can walk 0.15 of a walk(so confused lol)
A positive times a negative will always be negative
9 x -1 = -9
-9, Hope this helps!
C because it has an equal number of y's on each side so to solve it you would subtract 4y from both sides leaving:
5 = -6
which is not possible so there is no solutions.
Answer:
P(0) = 2000
P(10) = 2414
Step-by-step explanation:
The price for today is given by P(t) when t=0. Put 0 where t is, and do the arithmetic.
P(0) = 2000·1.019^0 = 2000
The price in 10 years is given by P(t) when t=10. Put 10 where t is, and do the arithmetic.
P(10) = 2000·1.019^10 ≈ 2000·1.207096 ≈ 2414
Prove we are to prove 4(coshx)^3 - 3(coshx) we are asked to prove 4(coshx)^3 - 3(coshx) to be equal to cosh 3x
= 4(e^x+e^(-x))^3/8 - 3(e^x+e^(-x))/2 = e^3x /2 +3e^x /2 + 3e^(-x) /2 + e^(-3x) /2 - 3(e^x+e^(-x))/2 = e^(3x) /2 + e^(-3x) /2 = cosh(3x) = LHS Since y = cosh x satisfies the equation if we replace the "2" with cosh3x, we require cosh 3x = 2 for the solution to work.
i.e. e^(3x)/2 + e^(-3x)/2 = 2
Setting e^(3x) = u, we have u^2 + 1 - 4u = 0
u = (4 + sqrt(12)) / 2 = 2 + sqrt(3), so x = ln((2+sqrt(3))/2) /3, Or u = (4 - sqrt(12)) / 2 = 2 - sqrt(3), so x = ln((2-sqrt(3))/2) /3,
Therefore, y = cosh x = e^(ln((2+sqrt(3))/2) /3) /2 + e^(-ln((2+sqrt(3))/2) /3) /2 = (2+sqrt(3))^(1/3) / 2 + (-2-sqrt(3))^(1/3) to be equ
= 4(e^x+e^(-x))^3/8 - 3(e^x+e^(-x))/2
= e^3x /2 +3e^x /2 + 3e^(-x) /2 + e^(-3x) /2 - 3(e^x+e^(-x))/2
= e^(3x) /2 + e^(-3x) /2
= cosh(3x)
= LHS
<span>Therefore, because y = cosh x satisfies the equation IF we replace the "2" with cosh3x, we require cosh 3x = 2 for the solution to work. </span>
i.e. e^(3x)/2 + e^(-3x)/2 = 2
Setting e^(3x) = u, we have u^2 + 1 - 4u = 0
u = (4 + sqrt(12)) / 2 = 2 + sqrt(3), so x = ln((2+sqrt(3))/2) /3,
Or u = (4 - sqrt(12)) / 2 = 2 - sqrt(3), so x = ln((2-sqrt(3))/2) /3,
Therefore, y = cosh x = e^(ln((2+sqrt(3))/2) /3) /2 + e^(-ln((2+sqrt(3))/2) /3) /2
= (2+sqrt(3))^(1/3) / 2 + (-2-sqrt(3))^(1/3)
