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CaHeK987 [17]
3 years ago
10

The total cost of hiring paul the plumber is a function of the required number of hours needed to complete the repair. Paul char

ges $40 per hour up to maximum of 8 hours, plus a $75 service charge what is the greatest value in the range for this situation
Mathematics
1 answer:
DiKsa [7]3 years ago
4 0

Paul surely charges 75 dollars, and then he adds 40 dollars per hours. This means that the expression for the money he asks is

m = 75+40h

where m is the amount of money and h is the number of hours.

Since h can be at most 8, this is the case that generates the highest money, so if we plug h=8 we have

m = 75+40\cdot 8 = 75 + 320 = 395

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Lady_Fox [76]

Answer:

1234588765

Step-by-step explanation:

5 0
3 years ago
Can some one really help me please!!!! :(((
GREYUIT [131]

Answer:

for the first on its a=24 and for the second its a= 48

Step-by-step explanation:

for the first one you have to find the length which is 8 and then you find the width which is 3 then you multiply that to get 24

for the second one you find the base which is the strait line which is 8 then you find the height which is 6 then you multiply it to get 48 i hopes this helped!

7 0
2 years ago
15pts please help asap<br> 2.5 as an improper fraction?
RideAnS [48]
2.5 = 2 5/10 as a Mixed fraction

To get a improper fraction, multiply the whole number with the denominator, and then add the numerator to the answer, giving the numerator (keep the denominator number)

2 x 10 = 20
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25/10 is your answer

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6 0
3 years ago
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Inessa05 [86]
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8 0
3 years ago
Which coefficient matrix represents a system of linear equations that has a unique solution ?
finlep [7]

Answer:

Option C

Step-by-step explanation:

We are given a coefficient matrix along and not the solution matrix

Since solution matrix is not given we cannot check for infinity solutions.

But we can check whether coefficient matrix is 0 or not

If coefficient matrix is zero, the system is inconsistent and hence no solution.

Option A)

|A|=\left[\begin{array}{ccc}4&2&6\\2&1&3\\-2&3&-4\end{array}\right] =0

since II row is a multiple of I row

Hence no solution or infinite

OPtion B

|B|=\left[\begin{array}{ccc}2&0&-2\\-7&1&5\\4&-2&0\end{array}\right] \\=2(10)-2(10)=0

Hence no solution or infinite

Option C

\left[\begin{array}{ccc}6&0&-2\\-2&0&6\\1&-2&0\end{array}\right] \\=2(36-2)=68

Hence there will be a unique solution

Option D

\left[\begin{array}{ccc}5&10&5\\4&1&4\\-1&-2&-1\end{array}\right] \\=2(10)-2(10)=0=0

(since I row is -5 times III row)

Hence there will be no or infinite solution

Option C is the correct answer



4 0
3 years ago
Read 2 more answers
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