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3 years ago

Where does the 1 come from in (1.0325)

Mathematics

1 answer:

NemiM [27]3 years ago

3 0

Answer: *1*.0325

Step-by-step explanation: When you round it to the nearest tenths place the 3 behind the 0 is less than 4 therefore everything behind the 0 gets taken away. The 0 has no value so you could take it away too along with the decimal point and it will turn into just 1.

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1. (3x3 + 3x2 - 4x + 5) + (x2 – 2x + x 4) |

KonstantinChe [14]

Answer:

$x^{2} + 3x^{3} + 4x^{2} - 6x + 5$

Step-by-step explanation:

first remove the parentheses

then add the like terms

lastly reorder the terms

6 0

3 years ago

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jeyben [28]

Answer:

3x3x12= volume of large prism= 108

1x1x12= volume of hole= 12

(Volume of prism)-(volume of hole)= 96

Step-by-step explanation:

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4 years ago

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What is the simplified form of -(x-4) ? 0 A -X-4 B. -X+ 0 C. X+4 D. X-4

sertanlavr [38]

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3 years ago

5a +4b+c-3a-c+4b simplified

MAVERICK [17]

Answer:

2a + 8b

Step-by-step explanation:

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3 years ago

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:

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

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

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

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

8 0

3 years ago