MATHEMATICS COLLEGE

Write the addition equation as a multiplication equation.
8 + 8 +8= 24

Answers

Answer 1

Answer:

Answer:

3 × 8 = 24

3 × 8 is the multiplication equation.

Answer 2

Answer:

Answer:

8x=24

Step-by-step explanation:

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Answers

Answer:

5)32

6)-7

7)-48

8)11

9)800

10)70.52

11)15

Step-by-step explanation:

Just as there are simultaneous algebraic equations (where a pair of numbers have to satisfy a pair of equations) there are systems of differential equations, (where a pair of functions have to satisfy a pair of differential equations).Indicate which pairs of functions satisfy this system. It will take some time to make all of the calculations.
y_1' = y_1 -2 y_2 \qquad y_2' = 3y_1 - 4 y_2

A. y_1 = \sin(x) +\cos(x) \qquad y_2 = \cos(x) - \sin(x)
B. y_1 = \sin(x) \qquad y_2 = \cos(x)
C. y_1 = \cos(x) \qquad y_2 = -\sin(x)
D. y_1 = e^{-x} \qquad y_2=e^{-x}
E. y_1 = e^x \qquad y_2=e^x
F. y_1 = e^{4x} \qquad y_2 = e^{4x}
G. y_1 = 2e^{-2x} \qquad y_2 = 3e^{-2x}

As you can see, finding all of the solutions, particularly of a system of equations, can be complicated and time consuming. It helps greatly if we study the structure of the family of solutions to the equations. Then if we find a few solutions we will be able to predict the rest of the solutions using the structure of the family of solutions.

Answers

Answer: D and G.

Step-by-step explanation:

For options D and G we will show that both differential equations are satisfied. For the other options we will show the pairs don't solve one of the equations.

A. $y_1 '= \cos(x)-\sin x$ and $y_1-2y_2= \sin x+\cos x -2(\cosx -\sin x )=3\sin x- \cos x \neq \cos x-\sin x$ (when x=0 the left side is -1 and the right side is 1) so the equation $y_1'=y_1 - 2y_2$ is not satisfied.

B. $y_2 '= -\sin x$ and $3y_1-4y_2= 3\sin x-4\cos x \neq -\sin x$ so the equation $y_2'=3y_1-4y_2$ is not satisfied.

C. $y_1 '= -\sin(x)$ and $y_1-2y_2= \cos x -2\sin x \neq -\sin x$ so these pairs don't solve the equation $y_1'=y_1-2y_2$ .

D. Since $y_1=y_2=e^(-x)$ then $y_1'=y_2'=-e^(-x)$ . The first equation is satisfied, because $y_1-2y_2=e^(-x)-2e^(-x)=-e^(-x)=y_1'$ . The second equation is also satisfied: $3y_1-4y_2=3e^(-x)-4e^(-x)=-e^(-x)=y_2'$ .

E. $y_2'=e^x$ and $3y_1-4y_2= 3e^x-4e^x=-e^x\neq -e^x$ so they don't satisfy the equation $y_2'=3y_1-4y_2$ .

F. $y_1 '= 4e^(4x)$ and $y_1-2y_2= e^(4x)-2e^(4x)=-e^(4x) \neq 4e^(4x)$ , then the equation $y_1'=y_1-2y_2$ is not satisfied.

G. In this case, $y_1=2e^(-2x)$ and $y_2=3e^(-2x)$ . Computing derivatives, $y_1'=-4e^(-2x)$ and $y_2'=-6e^(-2x)$ . The first equation is satisfied, because $y_1-2y_2=2e^(-2x)-6e^(-2x) =-4e^(-2x)=y_1'$ . The second equation is also satisfied: $3y_1-4y_2= 6e^(-2x)-12e^(-2x)=-6e^(-2x)=y_2'$ .

Final answer:

The pairs of functions that satisfy the given system of differential equations are Option D (y_1 = e^(-x), y_2 = e^(-x)) and Option E (y_1 = e^x, y_2 = e^x).

Explanation:

The given system of differential equations is:

y_1' = y_1 - 2y_2

y_2' = 3y_1 - 4y_2

To determine which pairs of functions satisfy this system, we can substitute each option into the system and check if they satisfy the equations.

Let's go through each option:

Option A: y_1 = sin(x) + cos(x), y_2 = cos(x) - sin(x)
By substituting these functions into the system, we get:
y_1' = cos(x) - sin(x) - 2(cos(x) - sin(x)) = -sin(x) - 4cos(x)
y_2' = sin(x) + cos(x) - 4(cos(x) - sin(x)) = 5sin(x) - 3cos(x)
These functions do not satisfy the system of differential equations.
Option B: y_1 = sin(x), y_2 = cos(x)
By substituting these functions into the system, we get:
y_1' = cos(x) - 2cos(x) = -cos(x)
y_2' = 3sin(x) - 4cos(x)
These functions do not satisfy the system of differential equations.
Option C: y_1 = cos(x), y_2 = -sin(x)
By substituting these functions into the system, we get:
y_1' = -sin(x) + 2sin(x) = sin(x)
y_2' = 3cos(x) - 4(-sin(x)) = 3cos(x) + 4sin(x)
These functions do not satisfy the system of differential equations.
Option D: y_1 = e^(-x), y_2 = e^(-x)
By substituting these functions into the system, we get:
y_1' = -e^(-x) - 2e^(-x) = -3e^(-x)
y_2' = 3e^(-x) - 4e^(-x) = -e^(-x)
These functions satisfy the system of differential equations.
Option E: y_1 = e^x, y_2 = e^x
By substituting these functions into the system, we get:
y_1' = e^x - 2e^x = -e^x
y_2' = 3e^x - 4e^x = -e^x
These functions satisfy the system of differential equations.
Option F: y_1 = e^(4x), y_2 = e^(4x)
By substituting these functions into the system, we get:
y_1' = 4e^(4x) - 2e^(4x) = 2e^(4x)
y_2' = 3e^(4x) - 4e^(4x) = -e^(4x)
These functions do not satisfy the system of differential equations.
Option G: y_1 = 2e^(-2x), y_2 = 3e^(-2x)
By substituting these functions into the system, we get:
y_1' = -2e^(-2x) - 2(3e^(-2x)) = -8e^(-2x)
y_2' = 3(2e^(-2x)) - 4(3e^(-2x)) = -6e^(-2x)
These functions satisfy the system of differential equations.

Therefore, the pairs of functions that satisfy the given system of differential equations are Option D (y_1 = e^(-x), y_2 = e^(-x)) and Option E (y_1 = e^x, y_2 = e^x).

Learn more about Systems of Differential Equations here:

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Which statement best describes the relationship between the graphs of the two linear equations below?3
y=-2 +4
2
3x - 2y = -8
The lines are
parallel
5
The lines intersect
and are
perpendicular
The lines intersect
and are not
perpendicular
The lines are the
same

Answers

Answer:

Step-by-step explanation:

In year 3 it is expected that the total value of clothing sales will reach 32 million if the total value of ASCO sells Remains the Same as year 2what percentage of sales clothing account for in year three

Answers

Answer:

The sales account for year 3 is $[\frac{32\ \text{mn}-x\ \text{mn}}{x\ \text{mn}}* 100\%]$ .

Step-by-step explanation:

As the sales for year 2 is not provided, assume it is x million.

The total sales in year 3 is, 32 million.

Compute the sales account for year 3 as follows:

$\text{Percentage of sales for year 3}=\frac{\text{Total Sales in year 3}-\text{Total Sales in year 2}}{\text{Total Sales in year 2}}* 100\%$

$=\frac{32\ \text{mn}-x\ \text{mn}}{x\ \text{mn}}* 100\%$

On an exam, the mean score is 78 points, with a standard deviation of 6 points. Assuming normal distribution of the scores, approximately what percentage of students received more than 85?A.12% B.17% C.8% D.9% E.None of the above

Answers

Answer: A. 12%

Step-by-step explanation:-

Given : In an exam , Mean score : $\mu=78\text{ points}$

Standard deviation : $6\text{ points}$

Let X be a random variable that represents the scores of students.

We assume that the points are normally distributed.

Z-score : $z=(x-\mu)/(\sigma)$

For x = 85, we have

$z=(85-78)/(6)\approx1.17$

Then using standard normal distribution table, the probability that the students received more than 85 is given by :-

$P(x>85)=P(z>1.17)=1-P(z<1.17)\n\n=1-0.8789995=0.1210005\approx0.12=12\%$

Hence, the percentage of students received more than 85 =12%

Which pair of variables would most likely have a negative correlation?A. the time spent driving in a car and the number of miles driven

B. the height of a dog in inches and the number of ounces of food the dog eats per day

C. the number of sit-ups completed and the number of calories burned during a workout

D. the time spent reading a book and the number of pages remaining to be read in the book

Answers

The pair of variables that will most likely have a negative correlation are D. the time spent reading a book and the number of pages remaining to be read in the book.

Negative Correlation

This means that the variables involved move in the opposite direction from each other.
Means that as one variable increases, the other decreases and vice versa.

If you spend a longer time reading a book, you would read more pages which would reduce the number of pages left to be read. The time taken to read the book is therefore negatively correlated with the number of pages left.

In conclusion, option D is correct.

Find out more on negative correlation at brainly.com/question/2476038.

Answer: D

Step-by-step explanation:

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