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raketka [301]
3 years ago
5

What is the best estimate for 591.3 divided by 29

Mathematics
1 answer:
ioda3 years ago
6 0
Best estimate would be to round 591.3 to 600 and round 29 to 30. 600/30 = 20.
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A right rectangular prism is shown.
Svet_ta [14]
So what you are looking for is the hypotenuse of the triangle DCF. 

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BC and CF and CF create a right triangle. 

Using Pythagorean Theorem, CF is about 10.2

Now using Pythagorean Theorem again, 3^2 +10.2^2=113

The sqaureroot of that is 10.6. So 10.6 is your answer
3 0
3 years ago
2.The number of grams A of a certain radioactive substance present at time, in yearsfrom the present, t is given by the formulaA
professor190 [17]

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)}

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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=0.000585A=0.000585

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

4 0
10 months ago
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gizmo_the_mogwai [7]

Answer:

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Step-by-step explanation:

where the box ends on the left side is the 1st quartile.

4 0
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likoan [24]

Answer:

$325

Step-by-step explanation:

5 0
2 years ago
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To solve this, you first subtract 60 by 18 to get 42

You then divide this by 2 to get 21

So, it'll take 21 minutes

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