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3 years ago

Point C is the center of the circle. The measure of angle ACB is 12x – 7. Arc ADB measures 10x – 7. Find the value of x.

1 answer:

6 0

We are asked to solve for the value of x given that the center is the point C and the ∠ACB is 12x-7 while the given arc is equal to 10x-7. To solve this problem, we need to remember that the summation of angle ACB and angle ADB is equal to 360°. Hence, the solution is shown below:
∠ACB + ∠ADB = 360°
12X - 7 + 10X -7 = 360 , combine same term such as:
22x - 14 = 360 , transpose -14 to the right side such as:
22x = 360 + 14
22x = 374 , divide both sides bu 22
22x/ 22 = 374 / 22
x = 17

The answer to this problem is 17 and this is the value of x.

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What are two consecutive integers whose sum is -97

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Answer:

-48 and -49

Step-by-step explanation:

n + n + 1 = -97

2n + 1 = -97

2n + 1 - 1 = -97 - 1

2n = -98

2n/2 = -98/2

n = -49

-49 + 1 = -48

-49, -48

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3 years ago

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Answer:

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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

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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.

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4 years ago

5. |4| = _____<br> 6. |-2 1/2| = _____<br> 7. |-8.673| = _____<br> 8. |-1,834| = _____

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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