Answer:
x = 2 and x = 10
Step-by-step explanation:
(x+3)^2 = (x - 5)^2 + (x + 2)^2
x^2 + 6x + 9 = x^2 - 10x + 25 + x^2 + 4x + 4
x^2 + 6x + 9 =2x^2 - 6x + 29
x^2 - 12x + 20 = 0
(x - 10)(x - 2) = 0
x - 10 = 0; x = 10
x - 2 = 0; x = 2
P(t) = 40(2)^(kt)
<span>when t=10, (1990), N = 55 </span>
<span>55 = 40(2)^(10k) </span>
<span>1.25 = 2^(10k) </span>
<span>take the ln of both sides, hope you remember your log rules </span>
<span>10k = ln 1.25/ln 2 </span>
<span>10k = .32193 </span>
<span>k = .032193 </span>
<span>so P(t) = 40(2)^(.032193t) </span>
<span>in 2000, t = 20 </span>
<span>P(20) = 40(2)^(.032193(20)) </span>
<span>= 62.5 million </span>
<span>for the formula </span>
<span>P(t) = a(2)^(t/d), d = the doubling time </span>
<span>so changing .032193t to t/d </span>
<span>= .032193t </span>
<span>= t/31.06 </span>
<span>so the doubling time is 31.06 </span>
<span>another way would be to set </span>
<span>80 = 40(2)^(.032193t) </span>
<span>2 = (2)^(.032193t) </span>
<span>.032193t = ln 2/ln 2 = 1 </span>
<span>t = 31.06</span>
Given :
A function, f(x) = -6 + 12
To Find :
The value of f(x) when x= -3, x= 0 and x=1.
Solution :
Given function is f(x) = -6 + 12 .
Simplifying above function, we get :
f(x) = 6
Now, the given function is independent of x.
So, for any value of x the the value of functions remains constant.
Hence, this is the required solution.