How many solutions does the equation have? 4(2n−4)+3=8n−19

Mathematics

2 answers:

kondaur [170]3 years ago

6 0

Are there answers? I would love to help more :)

san4es73 [151]3 years ago

6 0

No solution, because 4(2n-4)+3=8n-19 —> 8n-13=8n-19 —> -13=-19 and since there is no “n” then there is no answer/solution

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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,