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FromTheMoon [43]
3 years ago
7

Please answer this question! 20 points and brainliest!

Mathematics
1 answer:
azamat3 years ago
5 0

Step-by-step explanation:

You have to substitute... 0, 5, & 10, & 3

y = 2x + 3

y = 2(0) + 3

y = 0 + 3

y = 3

(0,3)

y = 2x + 3

y = 2(5) + 3

y = 10 + 3

y = 13

(5,13)

y = 2x + 3

y = 2(10) + 3

y = 20 + 3

y = 23

(10,23)

y = 2x + 3

y = 2(3) + 3

y = 6 + 3

y = 9

(3,9)





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According to one cosmological theory, there were equal amounts of the two uranium isotopes 235U and 238U at the creation of the
FromTheMoon [43]

Answer:

6 billion years.

Step-by-step explanation:

According to the decay law, the amount of the radioactive substance that decays is proportional to each instant to the amount of substance present. Let P(t) be the amount of ^{235}U and Q(t) be the amount of ^{238}U after t years.

Then, we obtain two differential equations

                               \frac{dP}{dt} = -k_1P \quad \frac{dQ}{dt} = -k_2Q

where k_1 and k_2 are proportionality constants and the minus signs denotes decay.

Rearranging terms in the equations gives

                             \frac{dP}{P} = -k_1dt \quad \frac{dQ}{Q} = -k_2dt

Now, the variables are separated, P and Q appear only on the left, and t appears only on the right, so that we can integrate both sides.

                         \int \frac{dP}{P} = -k_1 \int dt \quad \int \frac{dQ}{Q} = -k_2\int dt

which yields

                      \ln |P| = -k_1t + c_1 \quad \ln |Q| = -k_2t + c_2,

where c_1 and c_2 are constants of integration.

By taking exponents, we obtain

                     e^{\ln |P|} = e^{-k_1t + c_1}  \quad e^{\ln |Q|} = e^{-k_12t + c_2}

Hence,

                            P  = C_1e^{-k_1t} \quad Q  = C_2e^{-k_2t},

where C_1 := \pm e^{c_1} and C_2 := \pm e^{c_2}.

Since the amounts of the uranium isotopes were the same initially, we obtain the initial condition

                                 P(0) = Q(0) = C

Substituting 0 for P in the general solution gives

                         C = P(0) = C_1 e^0 \implies C= C_1

Similarly, we obtain C = C_2 and

                                P  = Ce^{-k_1t} \quad Q  = Ce^{-k_2t}

The relation between the decay constant k and the half-life is given by

                                            \tau = \frac{\ln 2}{k}

We can use this fact to determine the numeric values of the decay constants k_1 and k_2. Thus,

                     4.51 \times 10^9 = \frac{\ln 2}{k_1} \implies k_1 = \frac{\ln 2}{4.51 \times 10^9}

and

                     7.10 \times 10^8 = \frac{\ln 2}{k_2} \implies k_2 = \frac{\ln 2}{7.10 \times 10^8}

Therefore,

                              P  = Ce^{-\frac{\ln 2}{4.51 \times 10^9}t} \quad Q  = Ce^{-k_2 = \frac{\ln 2}{7.10 \times 10^8}t}

We have that

                                          \frac{P(t)}{Q(t)} = 137.7

Hence,

                                   \frac{Ce^{-\frac{\ln 2}{4.51 \times 10^9}t} }{Ce^{-k_2 = \frac{\ln 2}{7.10 \times 10^8}t}} = 137.7

Solving for t yields t \approx 6 \times 10^9, which means that the age of the  universe is about 6 billion years.

5 0
3 years ago
Select Is a Function or Is not a Function to correctly classify each relation.
Lunna [17]
<span>{(3, 7),(3, 6),(5, 4),(4, 7)}not
{(1, 5),(3, 5),(4, 6),(6, 4)}is
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{(0, 4),(3, 2),(4, 2),(6, 5)}is</span>
8 0
3 years ago
What is the <img src="https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B7%7D" id="TexFormula1" title="\sqrt[3]{7}" alt="\sqrt[3]{7}" alig
zvonat [6]

The answer you want will be 1.91293\dots

4 0
3 years ago
Find the common ratio for the geometric sequence defined by the formula: an=40(2‾√)n−1 a n = 40 ( 2 ) n − 1
WITCHER [35]

The ratio of the geometric sequence 402^{n-1} is 2.

Given that geometric sequence is 40*2^{n-1} and we have to find the common ratio of all the terms.

Geometric sequence is a sequence in which all the terms have a common ratio.

Nth termof a GP is ar^{n-1} in which a is first term and r is common ratio.

Geometric sequence=40*2^{n-1}

We have to first find the first term, second term and third term of a geometric progression.

First term=40*2^{1-1}

=40*2^{0}

=40*1

=40

Second term=40*2^{2-1}

=40*2^{1}

=40*2

=80

Third term=40*2^{3-1}

=40*2^{2}

=40*4

=160

Ratio of first two terms=80/40=2

Ratio of next two terms=160/80=2

Hence the common ratio of geometric sequence is 2.

Learn more about geometric progression at brainly.com/question/12006112

#SPJ1

4 0
2 years ago
If CDEF ≅ MNPQ, then EF≅ _____
Anuta_ua [19.1K]
Your answer would be PQ 
8 0
4 years ago
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