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3 years ago

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The inequality 4x + 260 < 400 represents the cost to repair Tom’s car. The cost of parts is $260 and it will take 4 hours of

work. If x represents the labor cost per hour and Tom needs to keep the total under $400, what is the highest labor cost per hour that he can afford?

1 answer:

madam [21]3 years ago

3 0

Answer: The highest labor cost per hour that he can afford is $34

Step-by-step explanation:

Hi, to answer this question we simply have to solve the inequality:

4x + 260 < 400

4x < 400-260

4x < 140

x < 140/4

x < 35

Since x represents the labor cost per hour , The highest labor cost per hour that he can afford is $34 (using whole numbers, because x must be less than 35, not less or equal)

Feel free to ask for more if needed or if you did not understand something.

You might be interested in

Can someone please help me with this

lana [24]

Answer:

3x-10+70°=180°(Supplementary angles)

3x+60°=180°

3x=180°-60°

3x=120°

3x/3 =120°/3

x=40°

Step-by-step explanation:

Hope that this is helpful.

Have a wonderful day.

6 0

2 years ago

Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______

Ludmilka [50]

Answer:

$(a)\ \sec^2(\theta) = 82$

$(b)\ \cot(\theta) = \frac{1}{9}$

$(c)\ \cot(\frac{\pi}{2} - \theta) = 9$

$(d)\ \csc^2(\theta) = \frac{82}{81}$

Step-by-step explanation:

Given

$\tan(\theta) = 9$

Required

Solve (a) to (d)

Using tan formula, we have:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

This gives:

$\frac{Opposite}{Adjacent} = 9$

Rewrite as:

$\frac{Opposite}{Adjacent} = \frac{9}{1}$

Using a unit ratio;

$Opposite = 9; Adjacent = 1$

Using Pythagoras theorem, we have:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

$Hypotenuse^2 = 9^2 + 1^2$

$Hypotenuse^2 = 81 + 1$

$Hypotenuse^2 = 82$

Take square roots of both sides

$Hypotenuse =\sqrt{82}$

So, we have:

$Opposite = 9; Adjacent = 1$

$Hypotenuse =\sqrt{82}$

Solving (a):

$\sec^2(\theta)$

This is calculated as:

$\sec^2(\theta) = (\sec(\theta))^2$

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

Where:

$\cos(\theta) = \frac{Adjacent}{Hypotenuse}$

$\cos(\theta) = \frac{1}{\sqrt{82}}$

So:

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

$\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2$

$\sec^2(\theta) = (\sqrt{82})^2$

$\sec^2(\theta) = 82$

Solving (b):

$\cot(\theta)$

This is calculated as:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

Where:

$\tan(\theta) = 9$ ---- given

So:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

$\cot(\theta) = \frac{1}{9}$

Solving (c):

$\cot(\frac{\pi}{2} - \theta)$

In trigonometry:

$\cot(\frac{\pi}{2} - \theta) = \tan(\theta)$

Hence:

$\cot(\frac{\pi}{2} - \theta) = 9$

Solving (d):

$\csc^2(\theta)$

This is calculated as:

$\csc^2(\theta) = (\csc(\theta))^2$

$\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2$

Where:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

$\sin(\theta) = \frac{9}{\sqrt{82}}$

So:

$\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2$

$\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2$

$\csc^2(\theta) = \frac{82}{81}$

4 0

3 years ago

Can any one help me out n this one pls

antoniya [11.8K]

It's D. 311 divided by 12 is 25.9

4 0

3 years ago

Read 2 more answers

Need help with question 2

yuradex [85]

1,5,8 would be the correct answer

8 0

3 years ago

Read 2 more answers

11 chocolate bars cost 10 dollars

inna [77]

I'm not sure what you're trying to ask, but if you're trying to find the individual cost;

Solution: Cost of 1 candy bar is $0.91

Explanation: The quotient of 10 and 11 is 0.90909090909, which rounds into 0.91

Hope This Helps!!!

-Austint1414

3 0

2 years ago