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4 years ago

6

- George wanted to buy a car for $20,000. He knew he wanted to trade it in when the car had depreciated to $12,500. He knows tha

t the car depreciates at a rate of 8.75%. Rounded to the nearest tenth of a year, how long would it be before he would have to trade in his car?

1 answer:

777dan777 [17]4 years ago

3 0

Answer:

yes your are correct socyeahhh

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If a phone card is usef to maje s long distance phone call yiu sre chsrged 0.75 per phone call plus an additional 0.20 per minut

Ksju [112]

Answer:

0.75+0.20m

Step-by-step explanation:

The start up price (0.75), plus the per min time, multiplied by the ammount of mins (m).

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3 years ago

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Find the sum. –7 + (–2)

Vika [28.1K]

Answer: -9

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3 years ago

What is the expected frequency of east campus and passed?

ser-zykov [4K]

The expected frequency of east campus and passed is C. 42 students

<h3>How to calculate the value?</h3>

The table for expected frequency is ,

East Campus West Campus Total

Passed (84*100)/22=42 (84*100)/200 =42 84

Failed (116*100)/200=58 (116*100)/22=58 116

Total 100 100 200

Passed = 84×100/200

= 42

Therefore, the expected frequency of East Campus and Passed is 42 students.

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3 0

1 year ago

Given: m∠V=103°, m∠VRT=71°, RS ∥ VU Find: m∠TRS, m∠U

lawyer [7]

Answer:

m∠U = 103° and m∠TRS = 6°

Step-by-step explanation:

In the given circle O,

Since, RS║VU, and VR is a transverse,

Therefor, m∠V + m∠R = 180° [Consecutive interior angles]

m∠R + 103° = 180° [m∠R = 103° given]

m∠R = 180° - 103°

m∠R = 77°

Since m∠R = m∠VRT + m∠TRS

77° = 71° + m∠TRS

m∠TRS = 77° - 71° = 6°

Quadrilateral RTUV is a cyclic quadrilateral.

Therefore, m∠U + m∠R = 180°

m∠U + 77° = 180°

m∠U = 180° - 77° = 103°

5 0

4 years ago

Ill give you Brainlyest! I need help ASAP work shown, please!!<br> problems 8-15

8_murik_8 [283]

Using the segment addition theorem, the solutions are:

5. x = 17

6. x = 7

7. x = 14

BC = 27

CD = 61

BD = 88

8. AB = 26

9. LJ = 46

10. x = 3

11. FG = 15

12. QS = 34

13. BC = 26

14. EG = 19

15. QS = 68

<h3>How to Apply the Segment Addition Theorem?</h3>

The segment addition theorem would be used to solve the problems as shown below:

5. UW = UV + VW [segment addition theorem]

Substitute

6x - 35 = 19 + 4x - 20

6x - 4x = 35 + 19 - 20

2x = 34

x = 17

6. HJ = HI + IJ [segment addition theorem]

Substitute

7x - 27 = 3x - 5 + x - 1

7x - 3x - x = 27 - 5 - 1

3x = 21

x = 7

7. BD = BC + CD [segment addition theorem]

Substitute

7x - 10 = 4x - 29 + 5x - 9

7x - 4x - 5x = 10 - 29 - 9

-2x = -28

x = 14

BC = 4x - 29 = 4(14) - 29 = 27

CD = 5x - 9 = 5(14) - 9 = 61

BD = 7x - 10 = 7(14) - 10 = 88

8. BC = BD

Substitute

2x + 1 = 5x - 26

2x - 5x = -1 - 26

-3x = -27

x = 9

AB = 43 - BC

AB = 43 - 2x + 1 = 43 - 2(9) + 1 = 26

9. 7x - 10 = 9x - 11 - (x + 3)

7x - 10 = 9x - 11 - x - 3

7x - 9x + x = 10 - 11 - 3

-x = -4

x = 4

LJ = 28 + 7x - 10 = 28 + 7(4) - 10

LJ = 46

10. 8x + 11 = 12x - 1

8x - 12x = -11 - 1

-4x = -12

x = 3

11. 11x - 7 = 3x + 9

11x - 3x = 7 + 9

8x = 16

x = 2

FG = 11x - 7 = 11(2) - 7

FG = 15

12. 5x - 3 = 21 - x

5x + x = 21 + 3

6x = 24

x = 4

QS = 2(5x - 3) = 10x - 6 = 10(4) - 6

QS = 34

13. 8x - 20 = 2(3x - 1)

8x - 20 = 6x - 2

8x - 6x = 20 - 2

2x = 18

x = 9

BC = AB = 3x - 1 = 3(9) - 1

BC = 26

14. 5x - 1 = 7x - 13

5x - 7x = 1 - 13

-2x = -12

x = 6

EG = 6x - 4 - 13 = 6(6) - 4 - 13

EG = 19

15. RT - ST = RS

Substitute

8x - 43 - (4x - 1) = 2x - 4

8x - 43 - 4x + 1 = 2x - 4

4x - 42 = 2x - 4

4x - 2x = 42 - 4

2x = 38

x = 19

QS = 2(RS)

QS = 2(2x - 4) = 4x - 8 = 4(19) - 8

QS = 68

Learn more about the segment addition theorem on:

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8 0

2 years ago