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AlexFokin [52]
3 years ago
5

Radioactive Decay Beginning with 16 grams of a radioactive element whose half -life is 30 years ,the mass y(in grams) remaining

after t years is given by
y=23(1/2)^t/45, t ≥ 0.
How much of the initial mass remains after 150 years?
Mathematics
1 answer:
Flura [38]3 years ago
5 0

Answer:

the equation should be corrected to fit the data of the problem.  With the corrected equation a mass of 0.5 grams remains after 150 years

Step-by-step explanation:

for the mass y( in grams)

y=23* (1/2)^(t/45), t ≥ 0.

the initial mass is at t=0 , then

y= 23 grams  → should be 16 grams

half-life from the equation = 45 years → should be 30 years

the correct equation should be

y=16*(1/2)^(t/30), t ≥ 0

then after 150 years  → t= 150

y=16*(1/2)^(150/30)= 16*(1/2)^5 = 16/32 = 0.5 grams

then a mass of 0.5 grams remains after 150 years

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xz_007 [3.2K]
Yeah, x= 7/2 and y= -5/2
Please mark as brainliest and I hope this helps
5 0
3 years ago
Can someone plz help me on this thank you and please stop giving me links
Brilliant_brown [7]

Answer: 900π

Step-by-step explanation: To calculate the area of a circle, we do πr^2, where r is the radius. The radius is 30, so the answer is 900π.

Hope this helps!

4 0
2 years ago
Prove the function f: R- {1} to R- {1} defined by f(x) = ((x+1)/(x-1))^3 is bijective.
Eduardwww [97]

Answer:

See explaination

Step-by-step explanation:

given f:R-\left \{ 1 \right \}\rightarrow R-\left \{ 1 \right \} defined by f(x)=\left ( \frac{x+1}{x-1} \right )^{3}

let f(x)=f(y)

\left ( \frac{x+1}{x-1} \right )^{3}=\left ( \frac{y+1}{y-1} \right )^{3}

taking cube roots on both sides , we get

\frac{x+1}{x-1} = \frac{y+1}{y-1}

\Rightarrow (x+1)(y-1)=(x-1)(y+1)

\Rightarrow xy-x+y-1=xy+x-y-1

\Rightarrow -x+y=x-y

\Rightarrow x+x=y+y

\Rightarrow 2x=2y

\Rightarrow x=y

Hence f is one - one

let y\in R, such that f(x)=\left ( \frac{x+1}{x-1} \right )^{3}=y

\Rightarrow \frac{x+1}{x-1} =\sqrt[3]{y}

\Rightarrow x+1=\sqrt[3]{y}\left ( x-1 \right )

\Rightarrow x+1=\sqrt[3]{y} x- \sqrt[3]{y}

\Rightarrow \sqrt[3]{y} x-x=1+ \sqrt[3]{y}

\Rightarrow x\left ( \sqrt[3]{y} -1 \right ) =1+ \sqrt[3]{y}

\Rightarrow x=\frac{\sqrt[3]{y}+1}{\sqrt[3]{y}-1}

for every y\in R-\left \{ 1 \right \}\exists x\in R-\left \{ 1 \right \} such that x=\frac{\sqrt[3]{y}+1}{\sqrt[3]{y}-1}

Hence f is onto

since f is both one -one and onto so it is a bijective

8 0
3 years ago
A container has
Aneli [31]

Answer:

there may be 16 but I am not sure

5 0
3 years ago
Which ordered pair in the form (a, b) is a solution of this equation?
Evgen [1.6K]

3a - 4b = 21


check (-2 , -3)


3(-2) -4(-3) =21


-6 +12  = 21


6 does not = 21  so NO


check (0 , 7)


3(0) -4(7) =21


0-28=21


-28 does not equal 21  so NO


check (-3 , -2)


3(-3) -4(-2) =21


-9 +8  = 21


1 does not = 21  so NO


check (7 , 0)


3(7) -4(0) =21


21  = 21

Choice D

5 0
3 years ago
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