8:00 a.m all the way until 2:00 p.m.
8:00 a.m. to 12:00 p.m is
4 hours.
12:00 p.m. to 2:00 p.m is
2 hours8:00 a.m. to
2:00 p.m. is
6 hours.
6 hours out of a 24 hour day.


The school holds classes for
25% of the 24-hour day.
Answer:
The rate of change will be $25
Step-by-step explanation:
Since she started from $45 we have to count again so on the chart it goes 5, 10, 15, 20, and 25!
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Answer:
3 days will pass
Step-by-step explanation:
work 24×1/3=8
sleep 24×3/8=9
meals 24×1/8=3
computer 24×1/6=4
reach up to 24 with sleep
9+9+9=27
which took a little over 3 days
Answer:
i) pi×4500 cm³
ii) pi×600 cm²
iii) 14 liters
Step-by-step explanation:
in general : the diameter is 30 cm, the radius is half of that (15 cm)
i)
the volume of a cylinder is base area times height.
Vc = pi×r²×h = pi×15²×20 = pi×225×20 = pi×4500 cm³
ii)
similar to volume, the side "mantle" area of the cylinder is the circumference of the base area times height.
surface area of the cylinder mantle is
Scm = 2×pi×r×h = 2×pi×15×20 = pi×30×20 = pi×600 cm²
iii)
for this we need now to do the multiplication with pi and then convert the cm³ to liters.
1 liter = a cube of 10 cm side length = 10×10×10 = 1000 cm³
pi×4500 = 14137.17 cm³ = 14.13717 liters or rounded 14 liters
Answer:
R3 <= 0.083
Step-by-step explanation:
f(x)=xlnx,
The derivatives are as follows:
f'(x)=1+lnx,
f"(x)=1/x,
f"'(x)=-1/x²
f^(4)(x)=2/x³
Simialrly;
f(1) = 0,
f'(1) = 1,
f"(1) = 1,
f"'(1) = -1,
f^(4)(1) = 2
As such;
T1 = f(1) + f'(1)(x-1)
T1 = 0+1(x-1)
T1 = x - 1
T2 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2
T2 = 0+1(x-1)+1(x-1)^2
T2 = x-1+(x²-2x+1)/2
T2 = x²/2 - 1/2
T3 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2+f"'(1)/6(x-1)^3
T3 = 0+1(x-1)+1/2(x-1)^2-1/6(x-1)^3
T3 = 1/6 (-x^3 + 6 x^2 - 3 x - 2)
Thus, T1(2) = 2 - 1
T1(2) = 1
T2 (2) = 2²/2 - 1/2
T2 (2) = 3/2
T2 (2) = 1.5
T3(2) = 1/6 (-2^3 + 6 *2^2 - 3 *2 - 2)
T3(2) = 4/3
T3(2) = 1.333
Since;
f(2) = 2 × ln(2)
f(2) = 2×0.693147 =
f(2) = 1.386294
Since;
f(2) >T3; it is significant to posit that T3 is an underestimate of f(2).
Then; we have, R3 <= | f^(4)(c)/(4!)(x-1)^4 |,
Since;
f^(4)(x)=2/x^3, we have, |f^(4)(c)| <= 2
Finally;
R3 <= |2/(4!)(2-1)^4|
R3 <= | 2 / 24× 1 |
R3 <= 1/12
R3 <= 0.083