Solve the following expression when t = 6 and d = 8 t + d = 8

The airplane in the scale drawing has a wingspan of a feet and a tailspan of b feet. if the tailspan in the drawing is 2 inches,

Elena L [17]

30 feet is correct. Since a = 3b, the wingspan is 6 inches in the drawing, so it is 6x5 = 30 feet on the plane.

7 0

3 years ago

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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

PLS help me if ure looking fotr branlyist

anzhelika [568]

Answer: $29

Step-by-step explanation: this is the easiest way to explain it-

20= 100%

10=50%

5=25%

10 is 50

9 is 45

8 is 40

7 is 35

6 is 30

5 is 25

3 0

3 years ago

How would i solve this? SOLVE FOR TU

creativ13 [48]

First subtract 30-23 that’d be the value of UV(7) now subtract 18 from 7, the answer will be 11 therefore TU is 11.

7 0

2 years ago

What is the domain and range of the relation shown in

Leya [2.2K]

Answer:

Domain : {10, 15, 19, 32}

Range: {-1, 5, 9}

Step-by-step explanation:

As the table of the relation is given as follows:

X Y

10 5

15 9

19 -1

32 5

From this table, we can define the relation by combining the ordered pairs from the table. Each and every order pair consists of x-coordinate and corresponding y-coordinate of any corresponding point.

So, the relation from the table can be made as follows:

relation : {(10, 5), (15, 9), (19, -1), (32, 5)}

Domain of a relation consists of all the x-coordinates (first elements) of order pairs.

Range of a relation consists of all the y-coordinates (second elements) of ordered pairs.

So, domain and range of relation will be as follows:

Domain : {10, 15, 19, 32}

Range: {-1, 5, 9}

Note: If there is any duplicate element in any x or y-coordinate of any ordered pair, it will be written only once when we determine domain and range. Here, in this example, 5 is duplicate, so, it will be mentioned only one time when we determine the range of this relation.

Keywords: domain, relation, range

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6 0

4 years ago

Solve the following expression when t = 6 and d = 8 t + d = 8 - 6