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NARA [144]
3 years ago
11

The half life of a certain substance is about 4 hours. The graph shows the decay of a 50 gram sample of the substance that is me

asured every hour for 9 hours.
Which function can be used to determine the approximate number of grams of the sample remaining after t hours?

a
y = 50(0.85)x

b
y = 25(0.15)x

c
y = 50(0.15)x

d
y = 25(0.85)x

Mathematics
2 answers:
olga55 [171]3 years ago
7 0
I can't see how any of those formulas show exponential decay.  (Did you type those correctly?)

The formula that will show the remaining amount correctly is:
ending amount = Bgng Amount / 2^n
where "n" is the number of haf-lives.

So, for example if half life = 4 hours and if we want to calculate the amount after 9 hours (that's 2.25 half-lives) then:
ending amount = Bgng Amount / 2^n
ending amount = 50 / (2^(9/4))
ending amount = 50 / 2^2.25
ending amount = 50 / <span><span><span>4.75682846 </span>
</span></span><span><span>ending amount = </span> 10.5112 grams
</span>Looking at the graph, we see that's about right.
 

Dimas [21]3 years ago
6 0

Answer:

a) y = 50(0.85)^x

Step-by-step explanation:

Let the function that shows the given situation is,

y=ab^x

Where a and b are any unknown numbers,

By the given diagram,

When x = 1, y = 42.5,

\implies 42.5 = ab^1

\implies ab = 42.5 ------(1)

Again, when x = 4, y = 26,

\implies 26 = ab^4

\implies 26 = ab(b^3)  

\implies 26 = 42.5(b^3)    ( By equation (1) )

\implies 0.611764706 = b^3

\implies 0.84890965425 = b

Again, By equation (1),

a = 50.0642203646

Hence, the equation that shows the given graph,

y = 50.0642203646(0.84890965425)^x

\implies y = 50(0.85)^x

⇒ Option a is correct.

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basically, we can use the vertical line test to see if it crosses at any 2 points

first one works
2nd one works
3rd works
4th works
5th works
they are all one to one functions



3.
easy
steps
for f(x)=?
replace f(x) with y
switch x and y
solve for y
replace y with  f⁻¹(x)
so

f(x)=(1+x)^3-2
replace
y=(1+x)^3-2
switch
x=(1+y)^3-2
solve
x+2=(1+y)^3
\sqrt[3]{x+2}=1+y
-1+\sqrt[3]{x+2}=
replace
f^{-1}(x)=-1+\sqrt[3]{x+2}
first option



4.
domain, hmm we can't divide by 0 so domain is all real numbers except for x=0
domain is (-\infty,0) \cup(0,\infty)
hmm, range
as x approaches negative or positive infinity, f(x)=5
as x approaches 0 from the left, f(x) appraoches negative infintiy
as x approaches 0 from the right , f(x) appraoches positive infintiy
so range is (-\infty,5) \cup(5,\infty)

domain is (-\infty,0) \cup(0,\infty)
range is (-\infty,5) \cup(5,\infty)
answer is the 4th option
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