Ask question

Ask question

All categories

3 years ago

5

In the number 1.1084, how does the value of the 1 in the ones place compare to the value of the 1 in the tenths place?

1 answer:

Vlada [557]3 years ago

3 0

Answer:The value of the 1 in the ones place is 10 times the value of the 1 in the tenths place.

Step-by-step explanation:

You might be interested in

40 PIONTS!!! HELP PLZ ANSWER THE QUESTIONS (IF YOU DONT ASWER THE QUETIONS THEN I WILL REPORT YOU)

allochka39001 [22]

Answer:

quadrant 1.

Step-by-step explanation:

A:(9,3)

B:(7,5)

C:(5,2)

Hope this helped you.

4 0

2 years ago

Can 57 rows be divided into equal sections

Serjik [45]

Yes it can be divided into 3 equal sections of 19

8 0

3 years ago

Read 2 more answers

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

Divide. (3 3/4 )÷(−2 1/2 ) Enter your answer as a mixed number, in simplified form, in the box.

Agata [3.3K]

Answer:

-11/14

Step-by-step explanation: first you divide 33/4 and (-21/2), then reduce the numbers to -33/4 times 2/21 after you reduce both fractions multiply -11/2 times 1/7 and you get -11/14 your alternative form would br -0.785714

8 0

3 years ago

Please help!!!! Consider the region shaded in yellow. Which inequality does the shaded region correspond to?

denis-greek [22]

Answer:

y-x-1 < 0

Step-by-step explanation:

according to the graph (0,1) and (-1,0) lies on the doted line,

equation of the dotted line is y-x-1=0

shaded region = y-x-1 < 0

5 0

1 year ago