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Answer in the attachment (graph C).
Answer:
x = -2
y = -1
(-2, -1)
General Formulas and Concepts:
<u>Pre-Algebra</u>
- Order of Operations: BPEMDAS
- Equality Properties
<u>Algebra I</u>
- Solving systems of equations using substitution/elimination
- Solving systems of equations by graphing
Step-by-step explanation:
<u>Step 1: Define systems</u>
y = x + 1
3x + 3y = -9
<u>Step 2: Solve for </u><em><u>x</u></em>
<em>Substitution</em>
- Substitute in <em>y</em>: 3x + 3(x + 1) = -9
- Distribute 3: 3x + 3x + 3 = -9
- Combine like terms: 6x + 3 = -9
- Isolate <em>x</em> term: 6x = -12
- Isolate <em>x</em>: x = -2
<u>Step 3: Solve for </u><em><u>y</u></em>
- Define original equation: y = x + 1
- Substitute in <em>x</em>: y = -2 + 1
- Add: y = -1
<u>Step 4: Graph systems</u>
<em>Check the solution set.</em>
If the radio active isotope has the half life of 24100 years, then the initial quantity is 2.16 grams
The half life in years = 24100
Consider the quantity of the radio active isotope remaining
y =
When t = 1000 the y = 1.2
y = C/2 when t = 1599
Substitute the values in the equation
C/2 =
Cancel the C in both side
1/2 =
Here we have to apply ln to eliminate the e terms
ln (1/2) = 24100k
k = ln(1/2) / 24100
k = -2.87× 10^-5
To find the initial value we have to substitute the value of k and y in the equation
1.2 = Ce^{1000 × -2.87× 10^-5}
C = 1.2 / e^(-0.0287)
C = 2.16 gram
Hence, the initial quantity of the radioactive isotope is 2.16 gram
Learn more about half life here
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