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sashaice [31]
3 years ago
11

50 points!!! Please help!

Mathematics
2 answers:
omeli [17]3 years ago
5 0

Each of the answers given at the right talks about the fourth month. So let's compare each of the fourth month's profits.


When the month t = 4 in Company B, the company made 10 hundred (or 1000) dollars in profits.


When the month t = 4 in Company A, we evaluate the function at t = 4. To do that we put t = 4 into the function.


P(4) = 1.8(1.4)⁴


P(4) = 6.91488


Thus the company made 691.488 dollars of profit. So during the 4th month, Company B made more than A.


Each of the answers also talks about year end profits, which would be after 12 months. The function for company B is linear whereas it is exponential for A. An exponential function will grow faster in A and have higher maximum values. We can conclude that year end profits for A will be higher.


We put the statements together - that B makes in the 4th month but at year's end A will make more.


Thus, the third box is the best answer.

mote1985 [20]3 years ago
4 0

Answer:

the third box.

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When adding or subtracting with scientific notation,why is it important to have the same power of 10?
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<h2>Explanation:</h2><h2></h2>

Because we perform something like common factor, so this guarantee that corresponding digits have regarding coefficients have the same place value. Remember that a number written in scientific notation is given by the product of a number between 1 and 10 and another number that is power of 10. For example, let's write two examples about this:

Addition:

a=2\times 10^3 \\ \\ b=4\times 10^3 \\ \\ \\ Then: \\ \\ a+b=2\times 10^3+4\times 10^3 \\ \\ =(2+4)\times 10^3 \\ \\ =5\times 10^3

Addition:

a=2\times 10^3 \\ \\ b=4\times 10^3 \\ \\ \\ Then: \\ \\ a+b=2\times 10^3+4\times 10^3 \\ \\ =(2+4)\times 10^3 \\ \\ =6\times 10^3

Subtraction:

a=7\times 10^5 \\ \\ b=2\times 10^5 \\ \\ \\ Then: \\ \\ a-b=7\times 10^5-2\times 10^5 \\ \\ =(7-2)\times 10^5 \\ \\ =5\times 10^5

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3 years ago
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.

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Simora [160]

Answer:

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What is the slope below
raketka [301]

Answer:

-1

Step-by-step explanation:

We can find the slope by change in y of change in x

change in y

---------------------

change in x

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the x changes +3 ( to the right 3)

-3

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