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2 years ago

11

George has 26 coins in quarters and dimes. He has a total of $4.70. How many quarters does he have?

1 answer:

Pachacha [2.7K]2 years ago

4 0

Answer:

he needs 19 quarters.

Step-by-step explanation:

1 quarter: 25 cents

19 quarters: $4.75

you need 19 quarters if you want have $4.75, but if it is with dimes and quarters then it is going to be 15 quarters and 10 dimes.

1 dimes: 10 cents

10 dimes: $1.00

1 quarter: 25 cents

15 quarters: $3.75

1.00+3.75=4.75

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Help me plz<br> thank u!

viktelen [127]

Answer:

x = 41

y = 65

z = 107

Step-by-step explanation:

y and 65 are vertically opposite angles. Vertically opposite angles are equal.

y = 65

74 + y + x = 180

74 + 65 + x = 180

139 + x = 180

x = 180 - 139

x = 41

z is the exterior angle, whereas 66 and x are interior angles in a triangle.

Formula : -

Exterior angle = Sum of two interior angles

z = 66 + x

= 66 + 41

z = 107

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2 years ago

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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.

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3 years ago

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Kobotan [32]

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