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3 years ago

925 round to the nearest hundred

2 answers:

4 0

925 rounded to the nearest hundred is 900 becuase first we identify the hundreds digit which in this case is 9. Second,we identify the next smallest place value (the digit to the right of the hundreds place) which in this case is 2. Is that digit greater than or equal to five? No - we round Down. The hundreds digit is stays the same but every digit after it becomes a zero.

vesna_86 [32]3 years ago

3 0

925 round to the nearest hundred

2 is after or before 5?
Before so we will round back, which would make the answer 900.

925 rounded to the nearest hundred = 900

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2 years ago

(x^2y+e^x)dx-x^2dy=0

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It looks like the differential equation is

$\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0$

Check for exactness:

$\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x$

As is, the DE is not exact, so let's try to find an integrating factor µ(x, y) such that

$\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0$

*is* exact. If this modified DE is exact, then

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}$

We have

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu$

Notice that if we let µ(x, y) = µ(x) be independent of y, then ∂µ/∂y = 0 and we can solve for µ :

$x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}$

The modified DE,

$\left(e^{-x}y + \dfrac1{x^2}\right) \,\mathrm dx - e^{-x}\,\mathrm dy = 0$

is now exact:

$\dfrac{\partial\left(e^{-x}y+\frac1{x^2}\right)}{\partial y} = e^{-x} \\\\ \dfrac{\partial\left(-e^{-x}\right)}{\partial x} = e^{-x}$

So we look for a solution of the form F(x, y) = C. This solution is such that

$\dfrac{\partial F}{\partial x} = e^{-x}y + \dfrac1{x^2} \\\\ \dfrac{\partial F}{\partial y} = e^{-x}$

Integrate both sides of the first condition with respect to x :

$F(x,y) = -e^{-x}y - \dfrac1x + g(y)$

Differentiate both sides of this with respect to y :

$\dfrac{\partial F}{\partial y} = -e^{-x}+\dfrac{\mathrm dg}{\mathrm dy} = e^{-x} \\\\ \implies \dfrac{\mathrm dg}{\mathrm dy} = 0 \implies g(y) = C$

Then the general solution to the DE is

$F(x,y) = \boxed{-e^{-x}y-\dfrac1x = C}$

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3 years ago

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Anettt [7]

Answer:

1) The size of the colony after 4 days is 6553 mosquitoes

2) After $t = 4.9\ days$

Step-by-step explanation:

To answer this question you must use the growth formula

$N = N_0e ^ {kt}$

Where

$N_0$ is the initial population of mosquitoes = 1000

t is the time in days

k is the growth rate

N is the population according to the number of days

We know that when t = 1 and $N_0=1000$ then N = 1600

Then we use these values to find k.

$1600 = 1000e ^{k(1)}\\\\\frac{1600}{1000} = e ^ k\\\\ln(\frac{1600}{1000}) = k\\\\k = 0.47$

Now that we know k we can find the size of the colony after 4 days.

$N = 1000e ^ {0.47(4)}\\\\N = 6553\ mosquitoes$

To know how long it should take for the population to reach 10,000 mosquitoes we must do N = 10000 and solve for t.

$10000 = 1000e ^ {0.47t}\\\\10 = e ^ {0.47t}\\\\ln(10) = 0.47t\\\\t = \frac{ln(10)}{0.47}\\\\t = 4.9\ days$

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2 years ago

Approximately how many inches are in 2500 millimeters? Given 1 in = 2.54 cm, 1 ft = .3048 m, 1 mi = 1.61 km.

love history [14]

Let's begin with 2500 mm and convert this to cm.

2500 mm = 250 cm

Next, convert 250 cm to inches. Recall that 1 inch = 2.54 cm.

Then (250 cm)(1 inch) / (2.54 cm) = (250 cm) (1 in)/ (2.54 cm)

= (250/2.54) inches = ? inches

7 0

3 years ago