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2 years ago

Drag each tile to the type of interest rate it describes

Mathematics

1 answer:

sertanlavr [38]2 years ago

6 0

Answer:

Nominal Interest Rate: Does not consider inflation provides a sense of purchasing power

Real Interest Rate: It describes the stated interest rate on an account and also considers inflation

Step-by-step explanation:

Real interest rate can be defined as the Nominal interest rate minus the inflation rate.

Nominal Interest Rate is the interest rate without taking into account the inflation

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14 divided by 2/7 how do I divide it and what is the anwer?

Anika [276]

Dividing by a fraction is like multiplying with the inverse.

So 14 / (2/7) is the same as 14 * 7/2, which is 49

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3 years ago

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Which is a correct first step in solving 5 – 2x < 8x – 3?

bezimeni [28]

Answer:

$\sf{5 < 10x - 3}$

Step-by-step explanation:

Hope it helps

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1 year ago

In the diagram below, a right circular cone with a radius of 3 inches has a slant height of 5 inches, and a right cylinder with

Westkost [7]

this is the correct answer

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3 years ago

Largest Possible GCF (Greatest Common Factor)

Ipatiy [6.2K]

Answer:

Step-by-step explanation:

To solve this problem we have to find the largest possible greatest common factor, which is 48.

Now, numbers 96 and 48 have as Greatest Common Factor 48, that's the largest number possible that is common to both numbers because

48/48 = 1

96/48 = 2

Therefore, the answer is 96 and 48, because they don't have repetitive digits.

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2 years ago

In each of Problems 5 through 10, verify that each given function is a solution of the differential equation.

WARRIOR [948]

Answer:

For First Solution: $y_1(t)=e^t$

$y_1(t)=e^t$ is the solution of equation y''-y=0.

For 2nd Solution: $y_2(t)=cosht$

$y_2(t)=cosht$ is the solution of equation y''-y=0.

Step-by-step explanation:

For First Solution: $y_1(t)=e^t$

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

$y_1(t)=e^t$

First order derivative:

$y'_1(t)=e^t$

2nd order Derivative:

$y''_1(t)=e^t$

Put Them in equation y''-y=0

e^t-e^t=0

0=0

Hence $y_1(t)=e^t$ is the solution of equation y''-y=0.

For 2nd Solution:

$y_2(t)=cosht$

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

$y_2(t)=cosht$

First order derivative:

$y'_2(t)=sinht$

2nd order Derivative:

$y''_2(t)=cosht$

Put Them in equation y''-y=0

cosht-cosht=0

0=0

Hence $y_2(t)=cosht$ is the solution of equation y''-y=0.

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3 years ago