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Delvig [45]
3 years ago
7

Can you put parenthesis inside parenthesis

Mathematics
2 answers:
stealth61 [152]3 years ago
7 0

Yes, you can put parenthesis inside of parenthesis.


Gnoma [55]3 years ago
4 0
Yes you can hope that helps
You might be interested in
Modeling Radioactive Decay In Exercise, complete the table for each radioactive isotope.
Julli [10]

Answer:

Step-by-step explanation:

Hello!

The complete table attached.

The following model allows you to predict the decade rate of a substance in a given period of time, i.e. the decomposition rate of a radioactive isotope is proportional to the initial amount of it given in a determined time:

y= C e^{kt}

Where:

y represents the amount of substance remaining after a determined period of time (t)

C is the initial amount of substance

k is the decaing constant

t is the amount of time (years)

In order to know the decay rate of a given radioactive substance you need to know it's half-life. Rembember, tha half-life of a radioactive isotope is the time it takes to reduce its mass to half its size, for example if you were yo have 2gr of a radioactive isotope, its half-life will be the time it takes for those to grams to reduce to 1 gram.

1)

For the first element you have the the following information:

²²⁶Ra (Radium)

Half-life 1599 years

Initial quantity 8 grams

Since we don't have the constant of decay (k) I'm going to calculate it using a initial quantity of one gram. We know that after 1599 years the initial gram of Ra will be reduced to 0.5 grams, using this information and the model:

y= C e^{kt}

0.5= 1 e^{k(1599)}

0.5= e^{k(1599)}

ln 0.5= k(1599)

\frac{1}{1599} ln 0.05 = k

k= -0.0004335

If the initial amount is C= 8 grams then after t=1599 you will have 4 grams:

y= C e^{kt}

y= 8 e^{(-0.0004355*1599)}

y= 4 grams

Now that we have the value of k for Radium we can calculate the remaining amount at t=1000 and t= 10000

t=1000

y= C e^{kt}

y_{t=1000}= 8 e^{(-0.0004355*1000)}

y_{t=1000}= 5.186 grams

t= 10000

y= C e^{kt}

y_{t=10000}= 8 e^{(-0.0004355*10000)}

y_{t=10000}= 0.103 gram

As you can see after 1000 years more of the initial quantity is left but after 10000 it is almost gone.

2)

¹⁴C (Carbon)

Half-life 5715

Initial quantity 5 grams

As before, the constant k is unknown so the first step is to calculate it using the data of the hald life with C= 1 gram

y= C e^{kt}

1/2= e^{k5715}

ln 1/2= k5715

\frac{1}{5715} ln1/2= k

k= -0.0001213

Now we can calculate the remaining mass of carbon after t= 1000 and t= 10000

t=1000

y= C e^{kt}

y_{t=1000}= 5 e^{(-0.0001213*1000)}

y_{t=1000}= 4.429 grams

t= 10000

y= C e^{kt}

y_{t=10000}= 5 e^{(-0.0001213*10000)}

y_{t=10000}= 1.487 grams

3)

This excersice is for the same element as 2)

¹⁴C (Carbon)

Half-life 5715

y_{t=10000}= 6 grams

But instead of the initial quantity, we have the data of the remaining mass after t= 10000 years. Since the half-life for this isotope is the same as before, we already know the value of the constant and can calculate the initial quantity C

y_{t=10000}= C e^{kt}

6= C e^{(-0.0001213*10000)}

C= \frac{6}{e^(-0.0001213*10000)}

C= 20.18 grams

Now we can calculate the remaining mass at t=1000

y_{t=1000}= 20.18 e^{(-0.0001213*1000)}

y_{t=1000}= 17.87 grams

4)

For this exercise we have the same element as in 1) so we already know the value of k and can calculate the initial quantity and the remaining mass at t= 10000

²²⁶Ra (Radium)

Half-life 1599 years

From 1) k= -0.0004335

y_{t=1000}= 0.7 gram

y_{t=1000}= C e^{kt}

0.7= C e^{(-0.0004335*1000)}

C= \frac{0.7}{e^(-0.0004335*1000)}

C= 1.0798 grams ≅ 1.08 grams

Now we can calculate the remaining mass at t=10000

y_{t=10000}= 1.08 e^{(-0.0001213*10000)}

y_{t=10000}= 0.32 gram

5)

The element is

²³⁹Pu (Plutonium)

Half-life 24100 years

Amount after 1000 y_{t=1000}= 2.4 grams

First step is to find out the decay constant (k) for ²³⁹Pu, as before I'll use an initial quantity of C= 1 gram and the half life of the element:

y= C e^{kt}

1/2= e^{k24100}

ln 1/2= k*24100

k= \frac{1}{24100} * ln 1/2

k= -0.00002876

Now we calculate the initial quantity using the given information

y_{t=1000}= C e^{kt}

2.4= C e^{( -0.00002876*1000)}

C= \frac{2.4}{e^( -0.00002876*1000)}

C=2.47 grams

And the remaining mass at t= 10000 is:

y_{t=10000}= C e^{kt}

y_{t=10000}= 2.47 * e^{( -0.00002876*10000)}

y_{t=10000}= 1.85 grams

6)

²³⁹Pu (Plutonium)

Half-life 24100 years

Amount after 10000 y_{t=10000}= 7.1 grams

From 5) k= -0.00002876

The initial quantity is:

y_{t=1000}= C e^{kt}

7.1= C e^{( -0.00002876*10000)}

C= \frac{7.1}{e^( -0.00002876*10000)}

C= 9.47 grams

And the remaining masss for t=1000 is:

y_{t=1000}= C e^{kt}

y_{t=1000}= 9.47 * e^{( -0.00002876*1000)}

y_{t=1000}= 9.20 grams

I hope it helps!

4 0
3 years ago
What does it mean for a ordered pair to be a solution to a linear equation​
jeka57 [31]

Answer:

What's a Solution to a System of Linear Equations? Note: If you have a system of equations that contains two equations with the same two unknown variables, then the solution to that system is the ordered pair that makes both equations true at the same time.

Step-by-step explanation:

8 0
3 years ago
Giving brainlist pls help Complete the table.
elena-s [515]

Answer:

(Since 3.14 seems more reliable, I'll use that for all four instead)

A: 17.6 ft

B: 81.6 mm

C: 9 cm

D: 12 in

Step-by-step explanation:

A:

2(2.8)*3.14\\5.6 * 3.14\\17.584 approx. = 17.6

B:

2(13)*3.14\\26 * 3.14\\81.64 approx. = 81.6

C:

\frac{56.5}{6.28}=\frac{2825}{314}=8\frac{313}{314}approx. = 9

D:

\frac{75.4}{6.28}=\frac{1885}{157}=12\frac{1}{157}approx. = 12

4 0
3 years ago
Find S5 for a geometric series for which a1=81 and r=1/9.
Minchanka [31]

ANSWER

S_5=91\frac{10}{81}




EXPLANATION


The sum of the first n terms of a geometric sequence is given by;


S_n=\frac{a_1(1-r^n)}{1-r} ,-1


Where n, is the number of terms and a_1 is the first term.


When n=5, we have a_1=81, we get;


S_5=\frac{81(1-(\frac{1}{9})^5)}{1-\frac{1}{9}}


S_5=\frac{81(1-\frac{1}{59049})}{1-\frac{1}{9}}


S_5=\frac{81(\frac{59048}{59049})}{\frac{8}{9}}




S_5=\frac{7381}{81}


S_5=91\frac{10}{81}




6 0
3 years ago
Brenda earns $1,700 per month after taxes. She is working on her budget and has the first three categories finished.
3241004551 [841]
Brenda earns $1,700 per month after taxes. She is working on her budget and has the first three categories finished.Housing $612 
Food $238 
Transportation $370 
What is the problem with this budget? 
The answer is => She has allotted more than 36% of her income for housing
8 0
3 years ago
Read 2 more answers
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