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3 years ago

Please HELP!!! Due tonight and I’m so worried.

Mathematics

2 answers:

adelina 88 [10]3 years ago

7 0

Answer:

3: Line AD

4: Line Segment AD

5: Point H

Step-by-step explanation:

For #3, because plane ABC and AEH are both planes that go in every direction infinitely, All points on line AD intersect.

For #4, because line segment AD is contained inside of the same plane of ABC, The entire line segment intersect.

For #5, Both lines meet at point H.

ExtremeBDS [4]3 years ago

6 0

Answer:

third one should be about f

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Which of the following expressions is correct?

Amiraneli [1.4K]

Answer:

The first one is correct

Step-by-step explanation:

Absolute value (Anything that is inside the tabs) will always be positive, even if there is a negative sign in the number. If the number is already positive, it's still positive.

4 0

3 years ago

Read 2 more answers

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

A. -78 <br> B. 49 <br> C. 53 <br> D. 14

____ [38]

Answer: 49

Step-by-step explanation:

5 0

2 years ago

3. The box-and-whisker plot below shows the heights,in inches, of the students in a 7th grade class.What percentage of the heigh

Fantom [35]

(3) 62.5% cuz u take the total number of the students and u divide it by the number of students between 60 to 65

7 0

3 years ago

Circle all of the equations

Tanzania [10]

Answer:

B and C

Step-by-step explanation:

Just not linear lol

3 0

3 years ago