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Ira Lisetskai [31]
3 years ago
6

Please I need help !!!!!

Mathematics
1 answer:
Mama L [17]3 years ago
4 0
The formula is y=0.5m+20

You start (0,20) then for each mile you increase by 0.5.
Or
Since the x axis is by two, then every two miles you increase by 1 mile. So (0,20); (2,21); (4,22); (6,23); (8,24).

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
The graph of y = h(x) is a line segment joining the points (1, -5) and (9,1).
Stella [2.4K]

Answer:

Ok, i cant drag the endpoints of the segment, but i can tell you how to do it.

First, we know that h(x) joins the points (1, -5) and (9, 1), then h(x) is a line:

h(x) = s*x + b

First, for a line that goes through the points (x1, y1) and (x2, y2), the slope will be:

s = (y2 -y1)/(x2 - x1)

Then in this case, the slope is:

s = (1 - (-5))/(9 - 1) = 0.75

Then we have

h(x) = 0.75*x + b

now, the value of b can be found as:

h(1) = -5 = 0.75*1 - b

b = - 5 - 0.75 = -5.75.

Then our equation is:

h(x) = 0.75*x - 5.75

Now, i gues you want to find the graph of:

y = h(-x)

Then our new function is:

g(x) = h(-x) =  -0.75*x - 5.75.

Now to find the points, we evaluate this function in the same values of x as before.

g(1) = -0.75*1 - 5,75 = -6,5

the point is (1, -6.5)

the second point is when x = 9.

g(9) = -0.75*9 - 5.75 = -12.5

The second point is (9, -12.5)

6 0
3 years ago
Read 2 more answers
Admission to a local aquarium is $10 for adults and S7 for children. The aquarium collected $6,459 from the sale of 783 tickets.
Anarel [89]

Answer:

The answer is 457

Step-by-step explanation:

You set up an equation as

10a+7c=6459

a+c=783

c= child

a= adult

then you simplify to 3a=978 by taking out one of the values.  You get a - 326.

783-326=457 so there is your answer

3 0
4 years ago
GIVING OUT BRAINLIST ASAP
rosijanka [135]

Answer:

$276.25

Step-by-step explanation:

325 * 0.85 = 276.25

5 0
3 years ago
Read 2 more answers
Evaluate 150 + 3p for p = 30
LiRa [457]

150+3p

p=30

150+3(30)

150+90

240

When p=30 in 150+3p, the final value is 240.

7 0
3 years ago
Read 2 more answers
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