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Tanya [424]
2 years ago
10

a clerk is paid $45.25 per hours for 40 hours a week, 1.50 times the regular rate of overtime and double the rate for a holiday.

How much does the clerk get if he works overtime for 5 hours and 2 hours on holidays?
Mathematics
1 answer:
KonstantinChe [14]2 years ago
7 0

Given :

A clerk is paid $45.25 per hours for 40 hours a week, 1.50 times the regular rate of overtime and double the rate for a holiday.

To Find :

How much does the clerk get if he works overtime for 5 hours and 2 hours on holidays.

Solution :

Amount from regular job = $ 45.25 × 40 = $1810 .

Amount from overtime = $ (45.25×1.5) × 5 = $339.375 .

Amount from holiday = $ (45.25×2) × 5 = $452.5 .

Total amount clerk will get is :

T = $( 1810 + 339.375 + 452.5 )

T = $2601.875

Hence, this is the required solution.

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Percents
Nana76 [90]

Answer:

As a percent: 115%

As a decimal: 1.15

As a fraction: \frac{23}{20}

Step-by-step explanation:

<u>Percentages</u>

Suppose Michael spends some amount of money on food, and that money is set at 100%. He leaves a 15% tip for the waitress.

Thus, he is really paying 100%+15% = 115% of the original price.

To express a percentage as a decimal, we just divide by 100:

115% = 115/100 = 1.15

The same division should be simplified instead of operated to express the percentage as a fraction:

115% = \frac{115}{100}

Simplifying by 5:

115% = \frac{23}{20}

Summarizing:

As a percent: 115%

As a decimal: 1.15

As a fraction: \frac{23}{20}

5 0
2 years ago
True or false. Tan^2 x = 1 - cos2x/ 1 + cos 2x
koban [17]

<u>ANSWER</u>

True

<u>EXPLANATION</u>

The given trigonometric equation is

\tan^{2} (x)  =  \frac{1 -  \cos(2x) }{1 +  \cos(2x) }

Recall the double angle identity:

\cos(2x)  =  \cos^{2} x -   \sin^{2}x

We apply this identity to obtain:

\tan^{2} (x)  =  \frac{1 - (\cos^{2} x -   \sin^{2}x) }{1 +  (\cos^{2} x -   \sin^{2}x) }

We maintain the LHS and simplify the RHS to see whether they are equal.

Expand the parenthesis

\tan^{2} (x)  =  \frac{1 - \cos^{2} x  +  \sin^{2}x }{1 +  \cos^{2} x -   \sin^{2}x}

\implies\tan^{2} (x)  =  \frac{1 - \cos^{2} x  +  \sin^{2}x }{1  -   \sin^{2}x  + \cos^{2} x }

Recall that:

1  -   \sin^{2}x  =  \cos^{2}x

1  -   \cos^{2}x  =  \sin^{2}x

We apply these identities to get:

\implies\tan^{2} (x)  =  \frac{\sin^{2}x +  \sin^{2}x }{\cos^{2} x + \cos^{2} x }

\implies\tan^{2} (x)  =  \frac{2\sin^{2}x }{ 2\cos^{2} x }

\implies\tan^{2} (x)  =  \frac{\sin^{2}x }{ \cos^{2} x }

\implies \tan^{2} (x)  =(  \frac{\sin x }{ \cos x })^{2}

Also

\frac{\sin x }{ \cos x } =  \tan(x)

\implies \tan^{2} (x)  =( \tan x )^{2}

\implies \tan^{2} (x)  =\tan^{2} (x)

Therefore the correct answer is True

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