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vovikov84 [41]
3 years ago
10

If you work 7 1/2 hours for 8 days and make 11 dollars an hour how much do you make at the end of the 8 days?

Mathematics
1 answer:
saw5 [17]3 years ago
5 0

Answer:

$660

Step-by-step explanation:

7.5 * 11 * 8 = $660

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astra-53 [7]
1/15 miles because you’re basically multiplying the fractions i believe
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What is the solution of the system? y=2 x= -1
kodGreya [7K]

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( -1,2)

Step-by-step explanation:

The solution is where the two lines intersect

The intersect at x=-1 and y=2

( -1,2)

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WILL GIVE BRAINLIEST, STARS, POINTS, AND THANK YOU FOR THE RIGHT ANSWER!!!
allochka39001 [22]

Answer:

1st blank: $51.00.    3rd blank: 1.1   5th blank: 24.2

2nd blank: $67.00.  4th blank: 91.30.  6th blank: -57.4

Step-by-step explanation:

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2 years ago
The profit P (in thousands of dollars) for an educational publisher can be modeled by P 52b31 5b21b where b is the number of wor
Anna007 [38]

Answer:

The lesser number of workbooks are 1,000

Step-by-step explanation:

The correct question is

The profit P (in thousands of dollars) for an educational publisher can be modeled by P=-b³+5b²+b where b is the number of workbooks printed (in thousands). Currently, the publisher prints 5000 workbooks and makes a profit of $5000. What lesser number of workbooks could the publisher print and still yield the same profit?

we have

P=-b^3+5b^2+b  

For P=\$5,000

substitute in the equation and solve for b

Remember that the profit and the number of workbooks is in thousands

so

P=5

5=-b^3+5b^2+b

Using a graphing tool

Solve the cubic function

The solutions are

x=-1

x=1

x=5

therefore

The lesser number of workbooks are 1,000

<u><em>Verify</em></u>

For b=1

P=-(1)^3+5(1)^2+1  

P=5  -----> is in thousands

so

P=\$5,000 ----> is ok

7 0
3 years ago
what is the smallest of 3 consecutive positive integers if the product of smaller two interfere is 5 less than 5 times the large
Lady_Fox [76]

Answer:

Two options:

5,6, and 7

-1,0,and 1

Step-by-step explanation:

Three consecutive natural numbers can be represented as n, n+1, and n+2.

The product of the smaller two would be n(n+1). The less than 5 times the larger is 5(n+2)-5.

Set them equal and solve by factoring:

n(n+1) = 5(n+2) -5\\n^2+n = 5n+10 -5\\n^2+n = 5n+5\\n^2 -4n-5=0\\(n-5)(n+1)=0

Set each factor equal to 0.

n-5==0 so n=5.

n+1=0 so n=-1.

This means the 3 consecutive numbers would be 5,6, and 7

OR

-1, 0 or 1.

8 0
3 years ago
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