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Yuliya22 [10]
2 years ago
12

You buy a pair of jeans at a department store. Jeans 39.99 Discount -10.00 Subtotal 29.99 Sales tax 1.95 Total 31.94 a. What is

the percent of discount to the nearest percent? The percent of discount is ___%. b. What is the percent of sales tax to the nearest tenth of a percent? The percent of sales tax is ___% c. The price of the jeans includes a 60% markup. After the discount, what is the percent of markup to the nearest percent? The percent of markup is ___%
Mathematics
1 answer:
Mariana [72]2 years ago
8 0
A)

if 39.99 is the 100%, what is 10 in percentage? well

\bf \begin{array}{ccllll}
amount&\%\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
39.99&100\\
10&x
\end{array}\implies \cfrac{39.99}{10}=\cfrac{100}{x}

solve for "x".

b)

now, with the discount, the amount is 29.99, thus if 29.99 is the 100%, what is 1.95 from it in percentage?

\bf \begin{array}{ccllll}
amount&\%\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
29.99&100\\
1.95&x
\end{array}\implies \cfrac{29.99}{1.95}=\cfrac{100}{x}

solve for "x".

c)

the original price is 39.99, the markup on that is 60%, how much is that?
well 

\bf \begin{array}{ccllll}
amount&\%\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
39.99&100\\
x&60
\end{array}\implies \cfrac{39.99}{x}=\cfrac{100}{60}\implies 39.99\cdot 60=100x
\\\\\\
\cfrac{39.99\cdot 60}{100}=x\implies 23.994=x

now, after the discount, the price is 29.99, how much is 23.994 in percentage of 29.99?

well  \bf \begin{array}{ccllll}
amount&\%\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
29.99&100\\
23.994&x
\end{array}\implies \cfrac{29.99}{23.994}=\cfrac{100}{x}

solve for "x".
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Answer:

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

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For 2nd Solution:y_2(t)=cosht

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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:

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2nd order Derivative:

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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.

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First order derivative:

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2nd order Derivative:

y''_2(t)=cosht

Put Them in equation y''-y=0

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

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Minato drove 390 miles. Part of the drive was along local roads, where his average speed was 20 mph, and the rest was along a hi
pashok25 [27]

Answer:

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Step-by-step explanation:

Given that the:

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Speed = distance/ time

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Along the highway,

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t = ( 390 - L)/60

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That is

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The LCM at right hand side will be 60

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