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3 years ago

Solve for x.

2 answers:

7 0

$-\dfrac{2}{3}(3x-4)+3x=\dfrac{5}{6}\ \ \ \ |\text{use distributive property}\\\\-\dfrac{2}{\not3_1}\cdot\not3^1x+\dfrac{8}{3}+3x=\dfrac{5}{6}\\\\-2x+3x+\dfrac{8}{3}=\dfrac{5}{6}\ \ \ \ |-\dfrac{8}{3}\\\\x=\dfrac{5}{6}-\dfrac{8}{3}\\\\x=\dfrac{5}{6}-\dfrac{8\cdot2}{3\cdot2}\\\\x=\dfrac{5}{6}-\dfrac{16}{6}\\\\\boxed{x=-\dfrac{11}{6}}$

Other method:

$-\dfrac{2}{3}(3x-4)+3x=\dfrac{5}{6}\ \ \ \ |\cdot6\\\\2(-2)(3x-4)+18x=5\\\\-4(3x-4)+18x=5\ \ \ \ |\text{use distributive property}\\\\-12x+16+18x=5\ \ \ \ |-16\\\\6x=-11\ \ \ \ |:6\\\\\boxed{x=-\dfrac{11}{6}}$

Vikentia [17]3 years ago

5 0

Answer:

x= -11/6

Step-by-step explanation:

I took the test!

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Factor the expression completely <br> 8y-36

Shtirlitz [24]

Answer:

4(2y - 9)

Step-by-step explanation:

$8y - 36 \\ = 4(2y - 9) \\$

5 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

Gabby is 2 kg heavier than her sister who weighs 3.4 kg. what is their combined weight?

lesya692 [45]

Answer:

8.8 kg

Step-by-step explanation:

yes

6 0

2 years ago

WILL GIVE BRAINLIEST

Paul [167]

256.
EXPLANATION:

When you are using exponents, you are basically just doing repeated multiplication.

Whatever the exponent (smaller number is) is how many times you multiply the base number by itself.

So 4 ^4 is 4 x 4 x 4 x 4. (4 fours.)

So 4 x 4 x 4 x 4 is 256.
HAVE A GUD DAY

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3 years ago

400 passengers go on a coach trip.

Nataliya [291]

Go be to if Derek ID do oh next 37

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3 years ago