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4 years ago

14

michaela has h hair ties. michaela's sister has triple the number of hair ties that michaela has. choose the expression that sho

ws how many hair bows michaela's sister has

1 answer:

kenny6666 [7]4 years ago

3 0

Answer:

Michaela = h

Michaela's sister = 3h

Answer = 3h

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puteri [66]

<span>1C. 5^3 = 125 mL
2A. 30^3 = 27,000 mL = 27 L
3D. 75 mm is 7.5 cm, so we do: 7.5^3 = 421.875 mL = 0.421875 L =~ 0.422 L

Note:
1 mL = 1 cm^3
1 L = 1,000 mL
10 mm = 1 cm</span>

5 0

3 years ago

Find the measure of the missing angle.

natulia [17]

Answer:

x = 76

Step-by-step explanation:

The inscribed angle x is half the measure of its intercepted arc.

x = $\frac{1}{2}$ × (102 + 50) = 0.5 × 152 = 76

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3 years ago

If (1/a)+(1/b)=10 what is the value of a+b?

Goshia [24]

B) 2/5

I think, if a and b are the same (coz then they would both equal 0.2)

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

4 years ago

Help me please on this math question

rjkz [21]

Answer:

729 * 10^-45

Step-by-step explanation:

Here, we want to expand the value based on the exponent

We have the power multiplying every term in the bracket

That would be;

9^3•10^-45

= 729 * 10^-45

5 0

3 years ago