Y = f(x) = cos(x)
= cos(180)
= -1
Part (i)
I'll use H in place of T to represent the heat of the object. That way there isn't a clash of variables lowercase t vs uppercase T.
The equation we're working with is updated to:
H(t) = 22 + a*2^(bt)
Plugging in t = 0 as the initial time value should lead to the temperature being H = 86 degrees Celsius.
So,
H(t) = 22 + a*2^(bt)
86 = 22 + a*2^(b*0)
86 = 22 + a*2^0
86 = 22 + a*1
86 = 22 + a
a+22 = 86
a = 86-22
a = 64
<h3>Answer: 64</h3>
=====================================================
Part (ii)
We'll use the value of 'a' we found earlier. Plus we'll use the fact that H = 28 when t = 0.5 (since 30 min = 30/60 = 0.5 hr).
H(t) = 22 + a*2^(bt)
28 = 22 + 64*2^(b*0.5)
28-22 = 64*2^(0.5b)
64*2^(0.5b) = 6
2^6*2^(0.5b) = 6
2^(6+0.5b) = 6
log( 2^(6+0.5b) ) = log(6)
(6+0.5b)*log(2) = log(6)
6+0.5b = log(6)/log(2)
6+0.5b = 2.5849625
0.5b = 2.5849625-6
0.5b = -3.4150375
b = -3.4150375/(0.5)
b = -6.830075
<h3>Answer: Approximately -6.830075</h3>
Answer:
10000000000
Step-by-step explanation:
10000000000
:) Hope this helps
