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allochka39001 [22]
2 years ago
8

• Jamie is constructing shipping containers for a sporting goods store. The containers are to be rectangular prisms.

Mathematics
1 answer:
Inga [223]2 years ago
4 0
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A conductor is mapping a trip and records the distance the train travels over certain time intervals.
leonid [27]
In half an hour the trains travels 22.5 miles
6 0
1 year ago
Find c1 and c2 such that M2+c1M+c2I2=0, where I2 is the identity 2×2 matrix and 0 is the zero matrix of appropriate dimension.
Katyanochek1 [597]

The question is missing parts. Here is the complete question.

Let M = \left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]. Find c_{1} and c_{2} such that M^{2}+c_{1}M+c_{2}I_{2}=0, where I_{2} is the identity 2x2 matrix and 0 is the zero matrix of appropriate dimension.

Answer: c_{1} = \frac{-16}{10}

             c_{2}=\frac{-214}{10}

Step-by-step explanation: Identity matrix is a sqaure matrix that has 1's along the main diagonal and 0 everywhere else. So, a 2x2 identity matrix is:

\left[\begin{array}{cc}1&0\\0&1\end{array}\right]

M^{2} = \left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]\left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]

M^{2}=\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]

Solving equation:

\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]+c_{1}\left[\begin{array}{cc}6&5\\-1&-4\end{array}\right] +c_{2}\left[\begin{array}{cc}1&0\\0&1\end{array}\right] =\left[\begin{array}{cc}0&0\\0&0\end{array}\right]

Multiplying a matrix and a scalar results in all the terms of the matrix multiplied by the scalar. You can only add matrices of the same dimensions.

So, the equation is:

\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]+\left[\begin{array}{cc}6c_{1}&5c_{1}\\-1c_{1}&-4c_{1}\end{array}\right] +\left[\begin{array}{cc}c_{2}&0\\0&c_{2}\end{array}\right] =\left[\begin{array}{cc}0&0\\0&0\end{array}\right]

And the system of equations is:

6c_{1}+c_{2} = -31\\-4c_{1}+c_{2} = -15

There are several methods to solve this system. One of them is to multiply the second equation to -1 and add both equations:

6c_{1}+c_{2} = -31\\(-1)*-4c_{1}+c_{2} = -15*(-1)

6c_{1}+c_{2} = -31\\4c_{1}-c_{2} = 15

10c_{1} = -16

c_{1} = \frac{-16}{10}

With c_{1}, substitute in one of the equations and find c_{2}:

6c_{1}+c_{2}=-31

c_{2}=-31-6(\frac{-16}{10} )

c_{2}=-31+(\frac{96}{10} )

c_{2}=\frac{-310+96}{10}

c_{2}=\frac{-214}{10}

<u>For the equation, </u>c_{1} = \frac{-16}{10}<u> and </u>c_{2}=\frac{-214}{10}<u />

6 0
1 year ago
What is the quotient (3x4 – 4x2 + 8x – 1) ÷ (x – 2) brainly
Karolina [17]

To find our quotient we are going to perform long division.

Long division step by step:

Step 1. Write the polynomial (dividend) in descending order (from highest to least exponent). If you encounter a missing term, use zero to fill the space of the term in the descending sequence.

Notice that even tough our polynomial is written in descending form, x^3 is missing, so we are going to use 0x^3 to fill the gap:

3x^4-4x^2+8x-1

3x^4+0x^3-4x^2+8x-1

Step 2. Divide the first term of the polynomial (dividend) by the first term of the divisor.

In our case the first term of the dividend is 3x^4 and the first term of the divisor is x, so \frac{3x^4}{x} =3x^3. Notice that 3x^3 is the first term of the quotient.

Step 3. Multiply the first term of the quotient by the terms of the divisor and subtract them from the respective term of the dividend.

The first term of our quotient (from the previous step) is 3x^3 and the divisor is (x-1), so 3x^3(x-2)=3x^4-6x^3. Now, we are going to subtract them from the respective term (the term with the same power) of the dividend. The respective terms of the dividend are 3x^4 and 0x^3, so 3x^4-3x^4=0 and 0x^3-(-6x^3)=0x^3+6x^3=6x^3

Step 4. Bring down the next term in the dividend and repeat the process for the remaining terms.

After finish the process (check the attached picture), we can conclude that the quotient of 3x^4-4x^2+8x-1 ÷ (x-2) is 3x^3+6x^2+8x+24 with a remainder of 47, or in a different notation: 3x^3+6x^2+8x+24+\frac{47}{x-2}

7 0
2 years ago
Read 2 more answers
Visa Card USA studied how frequently young consumers, ages 18 to 24, use plastic (debit and credit) cards in making purchases (A
umka2103 [35]

Answer:

Given the consumer is 18 to 24 years old, there is 37% probability that he uses a plastic card.

Step-by-step explanation:

This can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

P = \frac{P(B).P(A/B)}{P(A)}

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

In this problem, we have:

What is the probability of the consumer using a plastic card, given that the consumer is 18 to 24 years old.

The problem states that the probability that a consumer uses a plastic card when making a purchase is .37, so P(B) = 0.37

P(A/B) is the probability that the consumer being 18 to 24 years old, given that he uses a plastic card. The problem states that this probability is .19. So P(A/B) = 0.19

P(A) is the probability that the consumer is 18 to 24 years old. There is a .81 probability that the consumer is more than 24 years old. So there is a .19 probability that he is 18 to 24 years old. So P(A) = 0.19

The probability is

P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.37*0.19}{0.19} = 0.37

Given the consumer is 18 to 24 years old, there is 37% probability that he uses a plastic card.

4 0
2 years ago
Ashley invests $9,720 in a one-month money market account paying 3.16% simple annual interest and $8,140 in a two-year CD yieldi
tester [92]
For the first investment. A = P(1 + rt); where p = 9,720, r = 0.0316 and t = 1/12
A = 9720(1 + 0.0316/12) = 9720(1.0026) = $9,746

For the second investment,
A = 8140(1 + 0.0323 x 2) = 8140(1.0646) = $8,666

Total amount she had = $9,746 + $8,666 = $18,412
5 0
2 years ago
Read 2 more answers
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