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2 years ago

Find the lcm of 2.5, 1.0 and 70

Mathematics

1 answer:

Ivahew [28]2 years ago

4 0

Answer:

Step-by-step explanation:

2.5 = 5 * .5

1 = 1

70 = 2 * 5 * 7

LCM = 2 * 5 * 7

If you include the 1/2, you will reduce the LCM to 35, but 70 will be left out of the LCM.

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A given field mouse population satisfies the differential equation dp dt = 0.5p − 410 where p is the number of mice and t is the

ohaa [14]

Answer:

a) $t = 2 *ln(\frac{82}{5}) =5.595$

b) $t = 2 *ln(-\frac{820}{p_0 -820})$

c) $p_0 = 820-\frac{820}{e^6}$

Step-by-step explanation:

For this case we have the following differential equation:

$\frac{dp}{dt}=\frac{1}{2} (p-820)$

And if we rewrite the expression we got:

$\frac{dp}{p-820}= \frac{1}{2} dt$

If we integrate both sides we have:

$ln|P-820|= \frac{1}{2}t +c$

Using exponential on both sides we got:

$P= 820 + P_o e^{1/2t}$

Part a

For this case we know that p(0) = 770 so we have this:

$770 = 820 + P_o e^0$

$P_o = -50$

So then our model would be given by:

$P(t) = -50e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=-50 e^{1/2 t} +820$

$\frac{820}{50} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(\frac{82}{5}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(\frac{82}{5}) =5.595$

Part b

For this case we know that p(0) = p0 so we have this:

$p_0 = 820 + P_o e^0$

$P_o = p_0 -820$

So then our model would be given by:

$P(t) = (p_o -820)e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=(p_o -820)e^{1/2 t} +820$

$-\frac{820}{p_0 -820} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(-\frac{820}{p_0 -820})$

Part c

For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:

$12 = 2 *ln(\frac{820}{820-p_0})$

$6 = ln (\frac{820}{820-p_0})$

Using exponentials we got:

$e^6 = \frac{820}{820-p_0}$

$(820-p_0) e^6 = 820$

$820-p_0 = \frac{820}{e^6}$

$p_0 = 820-\frac{820}{e^6}$

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3 years ago

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Amiraneli [1.4K]

Answer:

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Step-by-step explanation:

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3 years ago

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jolli1 [7]

Answer:

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Step-by-step explanation:

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2 years ago

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bulgar [2K]

6 + 6 + 12 = 24, so how many atoms are in a molecule of glucose? That'd be 24.

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2 years ago

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Use summation notation to write the series -1 + 2 + 5 + 8 + 11... for 10 terms, then evaluate the sum.

garri49 [273]

Answer:

-1 + 2 + 5 + 8 + 11 + 14 + 17 + 20 + 23 + 26

10 terms

sum = 125

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3 years ago

Find the lcm of 2.5, 1.0 and 70​

Find the lcm of 2.5, 1.0 and 70