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Akimi4 [234]
3 years ago
5

7TH GRADE MATH!!! NEED ASAP

Mathematics
2 answers:
ludmilkaskok [199]3 years ago
8 0

Answer:

l >= 60

Step-by-step explanation:

l greater than or equal to 60

Bogdan [553]3 years ago
4 0

Answer:

l ≥ 60

Step-by-step explanation:

This means the l is greater than or equal to 60. In conclusion, the lightbulb is greater than or equal to 60 watts.

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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
Five friends all brought some cakes with them when they met. Each of them gave a cake to each of the others. They then ate all t
Sergeeva-Olga [200]

Answer:

1/2 i hope i helped

Step-by-step explanation:

8 0
3 years ago
Simplify (x-3/y^3)^-5
Katena32 [7]

Answer:

Correct simplification: y^15/(xy^3−3)^5

Step-by-step explanation:

4 0
2 years ago
What is 0.015 x 1.5=
alukav5142 [94]
The answer is 0.0225
7 0
3 years ago
Help plz nd thank you<br> (true or false questions)<br> 1.) -19-2+54&lt;12x97+-40<br> 2.) x+1=1+x
Anastaziya [24]
Number one is 33 < 1118 so true and number two is infinite solutions so it would be true as well
6 0
3 years ago
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