Answer:
c maybe 20
d maybe 33
and i dont know the last one
hola si esto está mal lo siento pero hice mi mejor esfuerzo y feliz gracias dando adiós que tengas un día bendito!
Let the distance be D miles
Car A:- D = 2.4R and for Car B:- D = 4(R- 22)
therefore 2.4R = 4(R-22)
4R - 88 = 2.4R
1.6R = 88
R = 55mph Answer
Answer:
yes
Step-by-step explanation:
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:
BRUUUUUUUUH
Step-by-step explanation:
