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zaharov [31]
3 years ago
14

Plz help... ANSWERS ONLY!

Mathematics
1 answer:
tekilochka [14]3 years ago
8 0

Answer:

1. 7 > 4 1/2           I'm only gonna do page one cause it's easy

2. 16 > 0

3. 3 1/2 < 5 1/2

Step-by-step explanation:

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F(prime)(t) = t^2 (1+f(t))<br><br> f(0) = 3<br><br> f(t) = ?
Andreyy89

Answer: F(t) = 4 e ^ {\frac{t^{3} }{3}}-1

Step-by-step explanation:

F'(t) = t^{2} (1+ F(t))

\frac{dF}{dt} = t^{2} (1+F)

First, we separated the variables:

\frac{dF}{1+F} = t^{2} dt

We integrate:

\int\limits^{F(t)}_{F(0)} {\frac{1}{1+F'} } \, dF' = \int\limits^t_0 {t'^{3}} \, dt

We change the variable for make more easy the integral:

u = 1+F', du = dF'

\int\limits^{}_{} {\frac{1}{u} } \, du = \int\limits^0_t {t'^{3}} \, dt

ln(u)= \frac{t^{3}}{3}

Now replace with u = 1+F' and evaluate the limit:

ln(1+F(t)) - ln (1+F(0)) = \frac{t^{3}}{3}

Using properties of natural logarithm

ln(1+F(t))= \frac{t^{3}}{3} + ln(4)

Taking exponential of both sides

e^{ln(1+F(t))} = e ^ {\frac{t^{3} }{3} + ln4}

e^{ln(1+F(t))} = e ^ {\frac{t^{3} }{3}}e^{ln4} =e ^ {\frac{t^{3} }{3}} * 4

1+F(t) = 4 e ^ {\frac{t^{3} }{3}}

F(t) = 4 e ^ {\frac{t^{3} }{3}}-1

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3 years ago
3438928 x 892 will mark brainlist
marin [14]

Answer:

3067523776

Step-by-step explanation:

hope this helped

5 0
3 years ago
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