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Cloud [144]
3 years ago
7

Luigi is an urban planner. as an independent contractor, he charges $140 fee plus $ 25 per hour for each contract with the city.

Mathematics
2 answers:
yaroslaw [1]3 years ago
6 0

x = hours 25 per hour = 25x

y = total

 so y = 25x +140

beks73 [17]3 years ago
6 0

Answer: y = 140+25x

Step-by-step explanation:

Since According to the question, Luigi initally charged $140.

And, per hour he charged an additional amount $25 for each contract with the city.

Here, x represents the number of hours worked on a project.

Thus the additional amount after x hours = 25x

So, the total amount he charged in x hours= 140+ 25x

And, according to the question y represents the above change.

Thus, we get a function for the given situtaion

which is , y= 140+ 25x , where x is number of hours and y is total amount charged.


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y = -4x+3

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Plywood comes in a variety of thickness for different uses put these in order from least to greatest
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Step-by-step explanation:

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3 years ago
An ordered pair of real numbers can be represented in a plane called the rectangular coordinate system or the ___________ plane
Ghella [55]
Coordinate plane is the answer
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Identify the functions that are continuous on the set of real numbers and arrange them in ascending order of their limits as x t
Studentka2010 [4]

Answer:

g(x)<j(x)<k(x)<f(x)<m(x)<h(x)

Step-by-step explanation:

1.f(x)=\frac{x^2+x-20}{x^2+4}

The denominator of f is defined for all real values of x

Therefore, the function is continuous on the set of real numbers

\lim_{x\rightarrow 5}\frac{x^2+x-20}{x^2+4}=\frac{25+5-20}{25+4}=\frac{10}{29}=0.345

3.h(x)=\frac{3x-5}{x^2-5x+7}

x^2-5x+7=0

It cannot be factorize .

Therefore, it has no real values for which it is not defined .

Hence, function h is defined for all real values.

\lim_{x\rightarrow 5}\frac{3x-5}{x^2-5x+7}=\frac{15-5}{25-25+7}=\frac{10}{7}=1.43

2.g(x)=\frac{x-17}{x^2+75}

The denominator of g is defined for all real values of x.

Therefore, the function g is continuous on the set of real numbers

\lim_{x\rightarrow 5}\frac{x-17}{x^2+75}=\frac{5-17}{25+75}=\frac{-12}{100}=-0.12

4.i(x)=\frac{x^2-9}{x-9}

x-9=0

x=9

The function i is not defined for x=9

Therefore, the function i is  not continuous on the set of real numbers.

5.j(x)=\frac{4x^2-7x-65}{x^2+10}

The denominator of j is defined for all real values of x.

Therefore, the function j is continuous on the set of real numbers.

\lim_{x\rightarrow 5}\frac{4x^2-7x-65}{x^2+10}=\frac{100-35-65}{25+10}=0

6.k(x)=\frac{x+1}{x^2+x+29}

x^2+x+29=0

It cannot be factorize .

Therefore, it has no real values for which it is not defined .

Hence, function k is defined for all real values.

\lim_{x\rightarrow 5}\frac{x+1}{x^2+x+29}=\frac{5+1}{25+5+29}=\frac{6}{59}=0.102

7.l(x)=\frac{5x-1}{x^2-9x+8}

x^2-9x+8=0

x^2-8x-x+8=0

x(x-8)-1(x-8)=0

(x-8)(x-1)=0

x=8,1

The function is not defined for x=8 and x=1

Hence, function l is not  defined for all real values.

8.m(x)=\frac{x^2+5x-24}{x^2+11}

The denominator of m is defined for all real values of x.

Therefore, the function m is continuous on the set of real numbers.

\lim_{x\rightarrow 5}\frac{x^2+5x-24}{x^2+11}=\frac{25+25-24}{25+11}=\frac{26}{36}=\frac{13}{18}=0.722

g(x)<j(x)<k(x)<f(x)<m(x)<h(x)

6 0
3 years ago
I need help with this!
lisabon 2012 [21]

Answer:

angle j = 90

angle k = 29

angle L = 61

Step-by-step explanation:

180 = 90 + (4x-19) + (5x+1)

90 = 9x - 18

108 = 9x

x = 12

3 0
3 years ago
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