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Setler [38]
3 years ago
6

Solve for x mathematics

Mathematics
2 answers:
Bas_tet [7]3 years ago
3 0

Answer:

x = 6

Step-by-step explanation:

9 / 6 = (2x - 6) / 4 (by the side splitter theorem)

6(2x - 6) = 36

12x - 36 = 36

12x= 72

x = 6

alexira [117]3 years ago
3 0
X=6 .................!
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What is the exact solution to the equation 4^5x=3^(x-2)<br> list the steps to get the answer
scoray [572]

Answer: x = -0.377

Step-by-step explanation:

We have the equation:

4^(5*x) = 3^(x - 2)

Now we can use the fact that:

Ln(A^x) = x*Ln(A)

Then we can apply Ln(.) to both sides of the equation to get:

Ln(4^(5*x)) = Ln(3^(x - 2))

(5*x)*Ln(4) = (x - 2)*Ln(3)

(5*x)*Ln(4) - x*Ln(3) = -2*Ln(3)

x*(5*Ln(4) - Ln(3)) = -2*Ln(3)

x = -2*Ln(3)/(5*Ln(4) - Ln(3)) = -0.377

6 0
3 years ago
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Answer:

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Step-by-step explanation:

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7 0
2 years ago
Penny joined a yoga studio that charges a $25 membership fee and $45 each month. Assume y = total cost and x = number of months.
nataly862011 [7]

Answer:

y=45x+25

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3 0
3 years ago
Write the prime factorization of the number. 18,234
shusha [124]

Answer:

18234=2\times 3\times 3\times 1013

Step-by-step explanation:

We are given that a number 18234

We have to find the prime factorization of the number

Prime factorization : The number written  is in  the product of prime numbers is called prime factorization.

In order to find the prime factorization we will find the factors of given number

18234=2\times 3\times 3\times 1013

Hence, the prime factorization of 18234=2\times 3\times 3\times 1013

3 0
3 years ago
The ​half-life of a radioactive element is 130​ days, but your sample will not be useful to you after​ 80% of the radioactive nu
gtnhenbr [62]

Answer:

We can use the sample about 42 days.

Step-by-step explanation:

Decay Equation:

\frac{dN}{dt}\propto -N

\Rightarrow \frac{dN}{dt} =-\lambda N

\Rightarrow \frac{dN}{N} =-\lambda dt

Integrating both sides

\int \frac{dN}{N} =\int\lambda dt

\Rightarrow ln|N|=-\lambda t+c

When t=0, N=N_0 = initial amount

\Rightarrow ln|N_0|=-\lambda .0+c

\Rightarrow c= ln|N_0|

\therefore ln|N|=-\lambda t+ln|N_0|

\Rightarrow ln|N|-ln|N_0|=-\lambda t

\Rightarrow ln|\frac{N}{N_0}|=-\lambda t.......(1)

                            \frac{N}{N_0}=e^{-\lambda t}.........(2)

Logarithm:

  • ln|\frac mn|= ln|m|-ln|n|
  • ln|ab|=ln|a|+ln|b|
  • ln|e^a|=a
  • ln|a|=b \Rightarrow a=e^b
  • ln|1|=0

130 days is the half-life of the given radioactive element.

For half life,

N=\frac12 N_0,  t=t_\frac12=130 days.

we plug all values in equation (1)

ln|\frac{\frac12N_0}{N_0}|=-\lambda \times 130

\rightarrow ln|\frac{\frac12}{1}|=-\lambda \times 130

\rightarrow ln|1|-ln|2|-ln|1|=-\lambda \times 130

\rightarrow -ln|2|=-\lambda \times 130

\rightarrow \lambda= \frac{-ln|2|}{-130}

\rightarrow \lambda= \frac{ln|2|}{130}

We need to find the time when the sample remains 80% of its original.

N=\frac{80}{100}N_0

\therefore ln|{\frac{\frac {80}{100}N_0}{N_0}|=-\frac{ln2}{130}t

\Rightarrow ln|{{\frac {80}{100}|=-\frac{ln2}{130}t

\Rightarrow ln|{{ {80}|-ln|{100}|=-\frac{ln2}{130}t

\Rightarrow t=\frac{ln|80|-ln|100|}{-\frac{ln|2|}{130}}

\Rightarrow t=\frac{(ln|80|-ln|100|)\times 130}{-{ln|2|}}

\Rightarrow t\approx 42

We can use the sample about 42 days.

7 0
3 years ago
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