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4 years ago

Find the value of the expression 2x^3+3y^2-17 when x=3 and y=4.

1 answer:

Crank4 years ago

8 0

2x^3+3y^2-17 if x=3 and y=4, so we have this:
2(3)^3+3(4)^2-17
First step, I will do the exponent first.
(3)^3=27
(4)^2=16, so
2(27)+3(16)-17
=54+48-17
=102-17
=85. Hope it help!

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Answer:

Question 1: $dm/dt=0.2m$

Question 2: $m=Ae^{(0.2t)}$

Question 3: $m=10e^{(0.2t)}$

Question 4: $m=10g$

Explanation:

Question 1: Write down the differential equation the mass of the bacteria, m, satisfies: m′= .2m

a) By definition: $m'=dm/dt$

b) Given: $rate=0.2m$

c) By substitution: $dm/dt=0.2m$

Question 2: Find the general solution of this equation. Use A as a constant of integration.

a) Separate variables

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b) Integrate

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c) Antilogarithm

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$m=e^{0.2t}\cdot e^C$

$e^C=A\\\\m=Ae^{(0.2t)}$

Question 3. Which particular solution matches the additional information?

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First, find the mass at t = 3 hours

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Now substitute in the general solution of the differential equation, to find A:

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Round A to 1 significant figure:

A = 10.

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Substitute t = 0 in the equation of the particular solution:

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Answer:

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