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3 years ago

A waitress earned $73 for 6 hours of work. The total included $46 in tips. What was her hourly wage? (Show Work) ...?

Mathematics

2 answers:

amid [387]3 years ago

6 0

Cross multiply dude money over money and x over hours

kipiarov [429]3 years ago

4 0

Since the waitress earned $73 together with the bonus, let's subtract it first to get the total wage for her 6 hours of work, so 73 - 46 = $27. After getting the total hourly wage, divide it at the number of hours worked which is 6 so <span>27 / 6 = $4.5 per hour.

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a gas station sells regular gas for $2.10 per gallon and premium gas for $2.90 a gallon. at the end of a business day 280 gallon

insens350 [35]

200 gallons of regular gas and 80 gallons of premium gas were sold.

Let the amount of regular gas sold be x gallons. So, the amount of premium gas sold will be (280 - x).

Now, forming the equation using the given information, to find the amount of each type of gallon.

Equation -

2.10x + 2.90(280 - x) = 652

Performing multiplication with values inside bracket in Left Hand Side

2.10x + 812 - 2.90x = 652

Rewriting the equation -

812 - 652 = 2.90x - 2.10x

Performing subtraction

160 = 0.8x

Rewriting the equation according to x

x = 160÷0.8

Performing division to find the value of x

x = 200

So, 200 gallons of regular gas was sold.

Amount of premium gas sold = 280 - 200

Performing subtraction

Amount of premium gas sold = 80 gallons

Hence, 200 gallons of regular gas and 80 gallons of premium gas was sold.

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1 year ago

In Ms. Perron's class, 75% of the studnetw are doys. There are 18boys in the class. What is the totla numder of students in Ms.

Anastasy [175]

Your answer should be 24

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3 years ago

If 180° < α < 270°, cos⁡ α = −817, 270° < β < 360°, and sin⁡ β = −45, what is cos⁡ (α + β)?

eduard

Answer:

$cos(\alpha+\beta)=-\frac{84}{85}$

Step-by-step explanation:

we know that

$cos(\alpha+\beta)=cos(\alpha)*cos(\beta)-sin(\alpha)*sin(\beta)$

Remember the identity

$cos^{2} (x)+sin^2(x)=1$

step 1

Find the value of $sin(\alpha)$

we have that

The angle alpha lie on the III Quadrant

The values of sine and cosine are negative

$cos(\alpha)=-\frac{8}{17}$

Find the value of sine

$cos^{2} (\alpha)+sin^2(\alpha)=1$

substitute

$(-\frac{8}{17})^{2}+sin^2(\alpha)=1$

$sin^2(\alpha)=1-\frac{64}{289}$

$sin^2(\alpha)=\frac{225}{289}$

$sin(\alpha)=-\frac{15}{17}$

step 2

Find the value of $cos(\beta)$

we have that

The angle beta lie on the IV Quadrant

The value of the cosine is positive and the value of the sine is negative

$sin(\beta)=-\frac{4}{5}$

Find the value of cosine

$cos^{2} (\beta)+sin^2(\beta)=1$

substitute

$(-\frac{4}{5})^{2}+cos^2(\beta)=1$

$cos^2(\beta)=1-\frac{16}{25}$

$cos^2(\beta)=\frac{9}{25}$

$cos(\beta)=\frac{3}{5}$

step 3

Find cos⁡ (α + β)

$cos(\alpha+\beta)=cos(\alpha)*cos(\beta)-sin(\alpha)*sin(\beta)$

we have

$cos(\alpha)=-\frac{8}{17}$

$sin(\alpha)=-\frac{15}{17}$

$sin(\beta)=-\frac{4}{5}$

$cos(\beta)=\frac{3}{5}$

substitute

$cos(\alpha+\beta)=-\frac{8}{17}*\frac{3}{5}-(-\frac{15}{17})*(-\frac{4}{5})$

$cos(\alpha+\beta)=-\frac{24}{85}-\frac{60}{85}$

$cos(\alpha+\beta)=-\frac{84}{85}$

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B.) would be the answer

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kondor19780726 [428]

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y=13+15x

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