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.

We know DC is 3cm, so we need CF.

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

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

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,