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2 years ago

Math help please Thanks A lot

Mathematics

1 answer:

babunello [35]2 years ago

8 0

Answer:

2 hours: 3968 [I don't understand the $ sign in the answer box]

At midnight: 12137

Step-by-step explanation:

The bacteria are increasing by 15% every hour. So for every hour we will have what we started with, plus 15% more.

The "15% more" can be represented mathematically with (1 + 0.15) or 1.15. Let's call this the "growth factor" and assign it the variable b. b is (1 + percent increase).

Since this per hour, in 1 hour we'll have (3000)*(1.15) = 3450

At the end of the second hour we're increased by 15% again:

(3450)*(1.15) = 3968.

Each additional hour add another (1.15) factor, If we assign a to be the starting population, this can be represented by:

P = a(1.15)^t for this sample that increase 15% per hour. t is time, in hours.

If a represents the growth factor, and P is the total population, the general expression is

P = ab^t

Using this for a = 3000 and b = 1.15, we can find the total population at midnight after starting at 2PM. That is a 10 hour period, so t = 10

P = (3000)*(1.15)^10

P = 12137

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irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

$y'(x)= x^2y(x)-\frac12y^2(x)$

Putting the value of y'(x) in equation (1)

$y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))$

Substituting x =0 and h= 0.2

$y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]$

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$\Rightarrow y(0.6)\approx 1.6593$

Substituting x =0.6 and h= 0.2

$y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]$

$\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]$

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Substituting x =0.8 and h= 0.2

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$\Rightarrow y(1.0)\approx 0.9152$

Therefore the value of y(1)= 0.9152.

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