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never [62]
3 years ago
10

Write an equation that represents the line. Use exact numbers.

Mathematics
2 answers:
Delicious77 [7]3 years ago
4 0

Answer:

4y+5x is ur equation for your linear line

photoshop1234 [79]3 years ago
4 0
4y+5x is your answer ❤️❗️✅❗️
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The correct code is CAHF.
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3 years ago
Read 2 more answers
the half-life of chromium-51 is 38 days. If the sample contained 510 grams. How much would remain after 1 year?​
madam [21]

Answer:

About 0.6548 grams will be remaining.  

Step-by-step explanation:

We can write an exponential function to model the situation. The standard exponential function is:

f(t)=a(r)^t

The original sample contained 510 grams. So, a = 510.

Each half-life, the amount decreases by half. So, r = 1/2.

For t, since one half-life occurs every 38 days, we can substitute t/38 for t, where t is the time in days.

Therefore, our function is:

\displaystyle f(t)=510\Big(\frac{1}{2}\Big)^{t/38}

One year has 365 days.

Therefore, the amount remaining after one year will be:

\displaystyle f(365)=510\Big(\frac{1}{2}\Big)^{365/38}\approx0.6548

About 0.6548 grams will be remaining.  

Alternatively, we can use the standard exponential growth/decay function modeled by:

f(t)=Ce^{kt}

The starting sample is 510. So, C = 510.

After one half-life (38 days), the remaining amount will be 255. Therefore:

255=510e^{38k}

Solving for k:

\displaystyle \frac{1}{2}=e^{38k}\Rightarrow k=\frac{1}{38}\ln\Big(\frac{1}{2}\Big)

Thus, our function is:

f(t)=510e^{t\ln(.5)/38}

Then after one year or 365 days, the amount remaining will be about:

f(365)=510e^{365\ln(.5)/38}\approx 0.6548

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2 years ago
How do I do solving systems of equations by substitution?
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Step 1:Solve one of the equations for either x = or y = .

Step 2:Substitute the solution from step 1 into the other equation.

Step 3:Solve this new equation.

Step 4:Solve for the second variable.

I hope this helps!

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O is the midpoint of Reason: Defintiin of a midpoint
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Step-by-step explanation:

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