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svetoff [14.1K]
3 years ago
9

Help please????????????

Mathematics
1 answer:
Aleks [24]3 years ago
7 0
The answer is b i think
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Colin buys a car for £15500.
timama [110]

Answer:

£14889.30

Step-by-step explanation:

The cost of the car = £15500.

For the first year,

depreciation = £15500 x 1%

                     = £155

Its worth after the first year = £15500 - £155

                             = £15345

For the 2nd year,

depreciation = £15345 x 1%

                     = £153.45

Its worth after the second year = £15345 - £153.45

                                 = £15191.55

For the 3rd year,

depreciation = £15191.55 x 1%

                      = £151.9155

Its worth after the third year = £15191.55 - £151.9155

                                            = £15039.6345

For the 4th year,

depreciation = £15039.6345 x 1%

                     = £150.3963

Its worth after the fourth year = £15039.6345 - £150.3963

                                             = £14889.2382

Thus, the worth of the car in 4 years would be £14889.30

8 0
2 years ago
What number is 75% of 8007
Sladkaya [172]

Answer:

600

Step-by-step explanation:

i hope tht helped I had tht question a couple days ago

6 0
3 years ago
Read 2 more answers
They both go together please help
Oxana [17]

Answer:

so you   got ehffffffffff

Step-by-step explanation:

7 0
2 years ago
Can someone help me solve these three problems please??
Tasya [4]
I'll solve 21, and you then should be able to solve the rest on your own!

Since ADC is 135, that means that that whole angle is 135 degrees. In addition, since angles ADB and BDC add up to ADC, we get ADB+BDC=ADC=135=11x+9+7x=18x+9. Subtracting 9 from both sides, we get 126=18x. Dividing both sides by 18, we get x=7. Plugging that into 11x+9=BDC, we get 11*7+9=77+9=86
4 0
3 years ago
Match each equation with its solution set.
taurus [48]

Answer:

1- The solution of I2x + 5I = 9 is {-7 , 2}

2- The solution of I2x + 7I + 2 = 11 is {-8 , 1}

3- The solution of I5 - xI = 6 is {-1 , 11}

4- The solution of I6x - 8I + 7 = 5 is ∅

5- The solution of Ix + 3I = 12 is {-15 , 9}

6- The solution of Ix - 3I = -12 is ∅

Step-by-step explanation:

* At first lets explain the meaning of IxI = a

- If IxI = a ⇒ then x = a or x = -a

- IxI never give a negative answer, because IxI means the

 magnitude of x is always positive

Ex: I-2I is 2

* Now lets find the solution of each equation

1- ∵ I2x + 5I = 9

∴ 2x + 5 = 9 ⇒ subtract 5 from both sides

∴ 2x = 4 ⇒ divide both sides by 2

∴ x = 2

OR

∴ 2x + 5 = -9 ⇒ subtract 5 from both sides

∴ 2x = -14 ⇒ divide both sides by 2

∴ x = -7

* The solution of I2x + 5I = 9 is {-7 , 2}

2- ∵ I2x + 7I + 2 = 11 ⇒ Subtract 2 from both sides

∴ I2x + 7I = 9

∴ 2x + 7 = 9 ⇒ subtract 7 from both sides

∴ 2x = 2 ⇒ divide both sides by 2

∴ x = 1

OR

∴ 2x + 7 = -9 ⇒ subtract 7 from both sides

∴ 2x = -16 ⇒ divide both sides by 2

∴ x = -8

* The solution of I2x + 7I + 2 = 11 is {-8 , 1}

3- ∵ I5 - xI = 6

∴ 5 -x = 6 ⇒ subtract 5 from both sides

∴ -x = 1 ⇒ divide both sides by -1

∴ x = -1

OR

∴ 5 -x = -6 ⇒ subtract 5 from both sides

∴ -x = -11 ⇒ divide both sides by -1

∴ x = 11

* The solution of I5 - xI = 6 is {-1 , 11}

4- ∵ I6x - 8I + 7 = 5 ⇒ Subtract 7 from both sides

∴ I6x - 8I = -2

- I  I never give negative answer

* The solution of I6x - 8I + 7 = 5 is ∅

5- ∵ Ix + 3I = 12

∴ x + 3 = 12 ⇒ subtract 3 from both sides

∴ x = 9

OR

∴ x + 3 = -12 ⇒ subtract 3 from both sides

∴ x = -15

* The solution of Ix + 3I = 12 is {-15 , 9}

6- ∵ Ix - 3I = -12

- I  I never give negative answer

* The solution of Ix - 3I = -12 is ∅

4 0
3 years ago
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