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Ahat [919]
3 years ago
7

A company used 7.4 x 105 sheets of paper a month, the accounting department used 8.9 x 103 sheets. How many were used by other d

epartments during the month
Mathematics
1 answer:
seraphim [82]3 years ago
4 0

Answer:

731,100

7.311 × 10^5

Step-by-step explanation:

The computation of the number of sheet used by other department is given below:

Given  that

The company used 7.4 × 10^5 sheets of paper So it would be 740,000  

And, The accounting department used 8.9 × 10^3 sheets so it would be 8,900  

Now the number of sheet used by other department is

= 740,000 - 8,900

= 731,100

= 7.311 × 10^5

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Solve this inequality : 5x+7x<36
Lera25 [3.4K]

Answer:

x < 3

Step-by-step explanation:

5x+7x<36

12x<36

Divide both sides by 12

x<3

7 0
3 years ago
Robbie was given $80 to spend at the Fair. His admission to the park costs $15.50 and each ride cost $5. He anticipates the cost
Ad libitum [116K]

Answer:

7 or 8

Step-by-step explanation:

Robbie was given $80.

His admission costs $15.50

80 - 15.50 = $64.50

He anticipates the cost of food at the fair will be $25.

64.50 - 25 = $39.50

What is the maximum number of rides he can take at the Fair?

(Total $ after admission and cost of food/ ride cost)

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4 0
3 years ago
Read 2 more answers
Question 4
Elanso [62]

Answer:

a A = 30 + 0.45 n

b A = 90.3

c 0 customers

Step-by-step explanation:

Given

Base\ Amount = 30

Rate = 0.45 per customer

Solving (a): Amount (A) on n customers

This is calculated as:

A = Base\ Amount + Rate * n

A = 30 + 0.45 * n

A = 30 + 0.45 n

Solving (b): Amount on 134 customers

In this case: n = 134

So:

A = 30 + 0.45 * 134

A = 30 + 60.3

A = 90.3

Solving (c): Customers at noon.

At noon, the amount is 30.

So:

A = 30 + 0.45 n

30 = 30 + 0.45n

Collect like terms

0.45n = 30 - 30

0.45n =0

n = 0

6 0
3 years ago
Sampson solves the inequality as shown below.
astraxan [27]

Answer:

subtract 3 from both sides

Step-by-step explanation:

7 0
3 years ago
Solve the system of equations.
guajiro [1.7K]

Answer:

Option 2: (1, 0) and (0, -5)

Step-by-step explanation:

Let's solve this system of equations using the elimination method.

Start by labelling the two equations.

5x -y= 5 -----(1)

5x² -y= 5 -----(2)

(2) -(1):

5x² -y -(5x -y)= 5 -5

Expand:

5x² -y -5x +y= 0

5x² -5x= 0

Factorise:

5x(x -1)= 0

5x= 0 or x -1= 0

x= 0 or x= 1

Now that we have found the x values, we can substitute them into either equations to solve for y.

Substitute into (1):

5(0) -y= 5 or 5(1) -y= 5

0 -y= 5 or -y= 5 -5

y= -5 or -y= 0

y= 0

Thus, the solutions are (0, -5) and (1, 0).

3 0
2 years ago
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