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3 years ago

Use row reduction to find the inverse of the given matrix if it exists, and check your answer by multiplication.

Mathematics

2 answers:

lakkis [162]3 years ago

8 0

Step-by-step explanation: see attachment below

Salsk061 [2.6K]3 years ago

7 0

Answer:

Step-by-step explanation:

Given the 2×2 matrix since the matrix consists of 2rows and 2columns, to find its inverse using row reduction method, we will augment the matrix given with a 2×2 identity matrix before carrying out the reduction on the resulting 2×4 matrices. The resulting matrix must be an identity matrix augmented with the inverse of the matrix in question. The process of reduction is as shown in simple steps.

[A | I] -> [I | A-¹] where

A is the given matrix

I is the 2×2 identity matrix

A-¹ is the inverse of the matrix

Check attachment for the reduced matrix.

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the half life of c14 is 5730 years. Suppose that wood found at an archeological excavation site contains about 35% as much C14 a

Furkat [3]

Answer:

The wood was cut approximately 8679 years ago.

Step-by-step explanation:

At first we assume that examination occured in 2020. The decay of radioactive isotopes are represented by the following ordinary differential equation:

$\frac{dm}{dt} = -\frac{m}{\tau}$ (Eq. 1)

Where:

$\frac{dm}{dt}$ - First derivative of mass in time, measured in miligrams per year.

$\tau$ - Time constant, measured in years.

$m$ - Mass of the radioactive isotope, measured in miligrams.

Now we obtain the solution of this differential equation:

$\int {\frac{dm}{m} } = -\frac{1}{\tau}\int dt$

$\ln m = -\frac{1}{\tau} + C$

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$ (Eq. 2)

Where:

$m_{o}$ - Initial mass of isotope, measured in miligrams.

$t$ - Time, measured in years.

And time is cleared within the equation:

$t = -\tau \cdot \ln \left[\frac{m(t)}{m_{o}} \right]$

Then, time constant can be found as a function of half-life:

$\tau = \frac{t_{1/2}}{\ln 2}$ (Eq. 3)

If we know that $t_{1/2} = 5730\,yr$ and $\frac{m(t)}{m_{o}} = 0.35$ , then:

$\tau = \frac{5730\,yr}{\ln 2}$

$\tau \approx 8266.643\,yr$

$t = -(8266.643\,yr)\cdot \ln 0.35$

$t \approx 8678.505\,yr$

The wood was cut approximately 8679 years ago.

5 0

3 years ago

Using all of the following numbers once, 8,8,3,5 how do u get 10

Sveta_85 [38]

Lol this is hard try using multiplication or division...

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3 years ago

Larry Mitchell invested part of gis 35,000 advance at 3% annual simple interest and the rest at 2% annual simple interest. If hi

Rom4ik [11]

Answer:

He invested $20000 for 3% rate and $15000 for 2% rate.

Step-by-step explanation:.

Let the 3% rate be for Account A and the 2% rate for Account B.

From the question, we know that the Principal from both accounts must add up to $35000

$P_A + P_B =$ 35000 ________________________ (1)

We also know that the interest from both accounts add up to $900

$I_A + I_B = $900$ ________________________(2)

The Interest from Account A (R = 3%, T = 1) is:

$I_A = \frac{P_A *R*T}{100}$

This implies that:

$I_A = \frac{P_A *3*1}{100}\\\\\\I_A = \frac{3P_A}{100} \\\\\\100*I_A = 3P_A\\\\\\100I_A - 3P_A = 0$ ________________________(3)

The Interest from Account B (R = 2%, T = 1) is:

$I_B = \frac{P_B *R*T}{100}$

This implies that:

$I_B = \frac{P_B *2*1}{100}\\\\\\I_B = \frac{2P_B}{100} \\\\\\100*I_B = 2P_B\\\\\\100I_B - 2P_B = 0$ _____________________________(4)

From (1),

$P_B = 35000 - P_A$

Putting this in (4)

$100I_B - 2*(35000 - P_A) = 0\\\\100I_B -70000 + 2P_A = 0\\\\100I_B +2P_A = 70000$ ________________(5)

From (2):

$I_A = 900 - I_B$

Putting this in (3):

$100(900 - I_B) - 3P_A = 0\\\\90000 - 100I_B - 3P_A = 0\\\\100I_B +3P_A = 90000$ _______________________(6)

(5) and (6) are simultaneous equations, hence, we can solve them:

$100I_B +3P_A = 90000\\\\100I_B + 2P_A = 70000$

Subtracting (5) from (6):

$3P_A - 2P_A = 90000 - 70000$

$P_A =$ $20000

Hence:

$P_B = 35000 - 20000\\\\$

$P_B =$ $15000

Also:

$100I_B + 3P_A = 90000\\\\100I_B + (3 * 20000) = 90000\\\\100I_B + 60000 = 90000\\\\\\100I_B = 90000 - 60000\\\\100I_B = 30000$

$I_B = 3000/100$ = $300

Hence:

$I_A = 900 - 300\\\\$

$I_A =$ $600

Hence, Larry invested $20000 at 3% annual interest and got $600 interest. He also invested $15000 at 2% annual interest and got $300.

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3 years ago

Stacy has hired Tom to build her a doll house. The doll house will be a scale model of her home, and the

Nikolay [14]

Answer:

56 in2

Step-by-step explanation:

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3 years ago

Which of these points lies on the line described by the equation below?

Pavlova-9 [17]

y – 5 = 6(x – 7)

y -5 = 6x - 42

y = 6x - 37

answer is B. (7, 5)

5 = 6(7) - 37

5 = 42 -37

5 = 5

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3 years ago