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KATRIN_1 [288]
3 years ago
11

Factorise x²-49please solve this answer​

Mathematics
2 answers:
velikii [3]3 years ago
8 0

Answer:

(x+7) (x-7)

Step-by-step explanation:

 \sqrt x^{2}-49

\sqrt x^{2} - \sqrt{49} = (x+7) (x-7)

Degger [83]3 years ago
4 0

Answer:

(x+7)(x-7)

Step-by-step explanation:

x²-49

=x²-7²

=(x+7)(x-7)

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
2 years ago
Stephen evaluated (6.34 x 10^-7)(4.5 x 10^3). His work is shown below. Which statements describe his errors? Check all that appl
Lostsunrise [7]
We have the following expression:
 (6.34 x 10 ^ -7) (4.5 x 10 ^ 3)
 The solution shown is:
 (6.34 x 10 ^ -7) (4.5 x 10 ^ 3)
 (6.34 x 4.5) (10 ^ -7 x 10 ^ 3)
 28.53 x 10 ^ -4
 -28.53 x 10 ^ 4
 -2.853 x 10 ^ 3
 The errors are:
 A. He changed the sign of the coefficient. A negative exponent does not affect the sign of a coefficient in scientific notation. The sign of the exponent determines the direction the decimal is moved in.
 D. He got the wrong value for the coefficients; 28.53 x 10^-4 is not possible. The coefficients in scientific notation are always greater than 1, but less than 10. 
3 0
3 years ago
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Just a follow up to my previous question "There is a 20% off sale going on at your local electronic store. The price of an mp3 p
Vikki [24]
The new price of the mp3 player is $44.95.
4 0
3 years ago
Order the numbers from least to greatest -3/4, -4/15, -2/5
barxatty [35]

Answer:

-3/4, -2/5, -4/15

Step-by-step explanation:

using calculator, -3/4 = -0.75, -4/15 = -0.266666..., -2/5 = -0.40

imagine that -0.75, -0.26... and -0.40 are temperatures

the coolest temperature is -0.75, the warmest one is -0.26...

so the least number is -0.75, greater -0.40 and the greatest -0.26...

4 0
3 years ago
Please please please help! I can Venmo you too if needed
Ugo [173]
Send it to a tutor !! on brainly
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3 years ago
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