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3 years ago

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The length of a rectangular vegetable bed is 3 feet more than twice the width. The length of the fence around the bed is 36 feet

. Find the length and with.
a) If the width of the rectangle is x, the equation is_.

b) The width of the vegetable bed is __, the length is __

1 answer:

Svetllana [295]3 years ago

8 0

B: The width of the vegetable bed is 5 feet. The length is 13 feet.

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Find a substance half-life in days. round your answer to the nearest tenths

jeka57 [31]

A 33 gram sample has k value of .1124
(Actually I think the problem SHOULD be worded: A 33 gram sample has k value of -.1124/days)

See attached graphic:
Half-Life = ln (.5) / k
Half-Life = -.693147 / -.1124/days
Half-Life = <span> <span> <span> 6.1667882562 </span> </span> </span> days

Source:
http://www.1728.org/halflif2.htm

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Answer:

Given that,

The number of grams A of a certain radioactive substance present at time, in years

from the present, t is given by the formula

$A=45e^{-0.0045(t)}$

a) To find the initial amount of this substance

At t=0, we get

$A=45e^{-0.0045(0)}$ $A=45e^0$

We know that e^0=1 ( anything to the power zero is 1)

we get,

$A=45$

The initial amount of the substance is 45 grams

b)To find thehalf-life of this substance

To find t when the substance becames half the amount.

A=45/2

Substitute this we get,

$\frac{45}{2}=45e^{-0.0045(t)}$

$\frac{1}{2}=e^{-0.0045(t)}$

Taking natural logarithm on both sides we get,

$\ln (\frac{1}{2})=-0.0045(t)^{}$ $(-1)\ln (\frac{1}{2})=0.0045(t)$ $\ln (\frac{1}{2})^{-1}=0.0045(t)$ $\ln (2)=0.0045(t)$ $0.6931=0.0045(t)$ $t=\frac{0.6931}{0.0045}$ $t=154.02$

Half-life of this substance is 154.02

c) To find the amount of substance will be present around in 2500 years

Put t=2500

we get,

$A=45e^{-0.0045(2500)}$ $A=45e^{-11.25}$ $A=45\times0.000013=0.000585$ $A=0.000585$

The amount of substance will be present around in 2500 years is 0.000585 grams

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Answer:

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