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slamgirl [31]
3 years ago
9

What is 9/5,-2.5,-1.1,-4/5,0.8 from least to greatest

Mathematics
1 answer:
padilas [110]3 years ago
7 0
Convert 9/5 and -4/5 to decimal form 
= 1.8 and -0.8

so answer is -2.5 , -1.1 . -0.8,  0.8, 1.8

= -2.5, -1.1 , -4/5, 0.8, 9/5
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Step-by-step explanation: because its going up 1/3 not down.


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How to solve the inequality
beks73 [17]

Answer:

x\geq 19/6

Step-by-step explanation:

1. Multiply both sides by 6 (the LCM of 6,3)

5/2 + 3-x <= 2(x-2)

2. Simplify 5/2+3-x to 11/2-x

11/2 - x<=2 (x-2)

3. Expand.

11/2-x<= 2x-4

4. Add x to both sides.

11/2<= 2x-4+x

5. Simplify 2x-4+x to 3x-4.

11/2<=3x-4

6. Add 4 to both sides.

11/2+4<=3x

7. Simplify 11/2+4 to 19/2

19/2<=3x

8. Divide both sides by 3

19/2 /3 <=x

9. Simplify 19/2 /3 to 19/2*3

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10. Simplify 2*3 to 6

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11. Switch sides.

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4 0
4 years ago
A basic cellular phone plan costs $4 per month for 70 calling minutes. Additional time costs $0.10 per minute. The formula C= 4+
Harrizon [31]

Answer:

For a monthly cost of at least $7 and at most $8, you can have between 100 and 110 calling minutes.

Step-by-step explanation:

The problem states that the monthly cost of a celular plan is modeled by the following function:

C(x) = 4 + 0.10(x-70)

In which C(x) is the monthly cost and x is the number of calling minutes.

How many calling minutes are needed for a monthly cost of at least $7?

This can be solved by the following inequality:

C(x) \geq 7

4 + 0.10(x - 70) \geq 7

4 + 0.10x - 7 \geq 7

0.10x \geq 10

x \geq \frac{10}{0.1}

x \geq 100

For a monthly cost of at least $7, you need to have at least 100 calling minutes.

How many calling minutes are needed for a monthly cost of at most 8:

C(x) \leq 8

4 + 0.10(x - 70) \leq 8

4 + 0.10x - 7 \leq 8

0.10x \leq 11

x \leq \frac{11}{0.1}

x \leq 110

For a monthly cost of at most $8, you need to have at most 110 calling minutes.

For a monthly cost of at least $7 and at most $8, you can have between 100 and 110 calling minutes.

8 0
3 years ago
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