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inessss [21]
2 years ago
8

63 first-graders and 21 other students attended a school assembly. What percentage of the

Mathematics
2 answers:
boyakko [2]2 years ago
4 0

Answer:

75%

Step-by-step explanation:

alexira [117]2 years ago
3 0

Answer:

75%

Step-by-step explanation:

First let's start by finding the total number of students at the assembly:

63+21=84

Then we find the percentage of the students that were first-graders.

63/84=9/12=3/4=75/100=75%

As further explanation to how this works:

We take the number of first graders over the total number of students so we can find the percentage. Then we simplify it down to 3/4. Since percentages are out of 100 we can change the fraction so it's something out of 100. We multiply the numerator and denominator by 25 (since 100/4=25) and we get 75/100 which is 75%

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Brad earns $12.75/hr and works 40 hours a week. What is his gross annual income?
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Step-by-step explanation:

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4 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

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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

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

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3 years ago
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