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Galina-37 [17]
2 years ago
15

Over six months, a family's electric bills averaged $55 per month. The bills for

Mathematics
1 answer:
kenny6666 [7]2 years ago
7 0
Answer:

$54.35

Step-by-step explanation:

The differences from the average value are;

2.60, 5.00, -1.75, -4.25, -0.95
The total of these differences is +0.65, so the final bill must have a value that is $0.65 lower than the average.

The bill for the 6th month is $54.35.
_____
Check
The average of the bills for 6 months is;

(57.60 +60 +53.25 +50.75 +54.05 +54.35)/6 = 330/6 = 55

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Is this Jose Galvan all you have to do is multiply 3.14by your circumference
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One eighth of forty eight
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One eighth of forty eight is six

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At her juice parlor, Mrs. Jones makes a different amount of fruit punch each day by mixing grape juice and peach juice. The tabl
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Step-by-step explanation:

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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

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Hence,

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

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Square the binomial<br>(2x - 3)²​
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Step-by-step explanation:

Change it into this

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Simplify again

4x^2 - 12x + 9

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