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aivan3 [116]
3 years ago
15

What expression is equal to (3x^2yz) (5xy^3z^2)

Mathematics
1 answer:
Svetlanka [38]3 years ago
8 0
Your Answer is <span>15x^3y^4z^3</span>
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Find x for all three triangles.
Lady bird [3.3K]
In 13, x=27.496
In 14, x=16.824
In 15, x=51.34° 

<em />
I did this in my head; it would take longer to write it out. I'll show you the trig if need be.
3 0
3 years ago
A pipe is leaking at 1.5 cups per day. About how many gallons per week is the pipe leaking
Soloha48 [4]
Multiply 1.5 times 7. That comes to 10.5 cups per week. Now convert it to quarts. There are 4 cups in one quart, so divide 10.5 by 4. That comes to 2.625 quarts per week. Now convert it to gallons. There are 4 quarts in a gallon, so divide 2.625 by 4. That comes to about 0.7 gallons a week, or 7/10.
8 0
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.

5 0
3 years ago
I need help As Soon As possible!
Julli [10]
This should be simple, add the Amounts of snow for the first week
4.32 + 6.86 = 11.18
-------------------------------------------------------
The subtract the amounts of snow from each week
11.18 - 7.89 = 3.29
---------------------------------------- 
C.) 3.29 is your answer
6 0
3 years ago
Elena's account balance with her parents is -5.50$ she adds a certain amount of money to her balance by mowing
8090 [49]

Answer:

big dawg

Step-by-step explanation:happy

4 0
3 years ago
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