MATHEMATICS COLLEGE

Simplify i^15. (imaginary numbers to the 15th power)

Answers

Answer 1

Answer: i x i x i x i x i x i x i x i x i x i x i x i x i x i x i

Answer 2

Answer:

Answer:

Its -i on edge

Step-by-step explanation:

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Nicole buys candy that costs $4 per pound. She will spend more than $28 on candy. What are the possible numbers of pounds she will buy?

Answers

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Given C=n+15, determine the value of C when n=4

Answers

Answer:

Step-by-step explanation:

If N=4 and the problem is C= N+15 that means you add 4 to 15 and then get 19 which will be your answer for C. You can also check this by subtracting 19-15 or 19-4!!! I hope this helps!!! Have a great day whom ever you are

An experiment is pulling a ball from an urn that contains 3 blue balls and 5 red balls. a.) Find the probability of getting a red ball. b.) Find the probability of getting a blue ball. c.) Find the odds for getting a red ball. d.) Find the odds for getting a blue ball.

Answers

Answer:

Step-by-step explanation:

The urn contains 3blue balls 5 red balls

a) probability of getting a red ball

P=no of favourable of outcomes /total no outcomes

P(red ball) = 5/8

b) Probability of blue ball

P(blue ball) = 3/8

c) Odds getting a red ball

odds in favour of any object = m/n

m : event to occur

n : event will not occur

Odds(red ball) = 5/3

Odds(blue) = 3/5

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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To make lemonade you can mix 4 teaspoons of lemonade powder with 16 ounces of water. What is the ratio of powder towater?
4:32
32:8
24:64
32:128

Answers

Answer:

32:128

Step-by-step explanation:

divide all of it by 2, you get 16:64. Again, 8:32. Again, 4:16

A builder needs to tile a 7-foot by8-foot room with one-foot square
tiles. How many tiles will she need?
56 tiles
15 tiles
30 tiles

Answers

Final answer:

To tile a 7-foot by 8-foot room with one-foot square tiles, the builder will need 56 tiles.

Explanation:

To find the number of tiles needed to tile a room, we can divide the area of the room by the area of each tile. The area of the room is 7 feet by 8 feet, which is 56 square feet. The area of each tile is 1 square foot. So, to find the number of tiles needed, we divide 56 by 1, which gives us 56 tiles.

Learn more about Calculating the number of tiles needed here:

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