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WITCHER [35]
3 years ago
5

A car sells for 16,000. If the rate of depreciation is 18%, find the value of the car after 8 years.

Mathematics
1 answer:
Black_prince [1.1K]3 years ago
6 0

Answer:

Step-by-step explanation:

price after 8 years Amount=P(1-0.18)^8

Amount=16,000(1-0.18)^8

=16,000(0.82)^8

≈3270.63 $

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Christian is planning a vacation and finds an airplane ticket that cost 180$. He found a promo code offering 17% off. Write a si
lesya [120]

Answer:

83/100 x 180

Step-by-step explanation:

We have to find the cost of the ticket after the discount.

100% - 17%= 83%

83% of the total percentage of the original price that he needs to pay.

So we do:

83/100 x 180

So, this would be the expression.

Hope this helps!:)

7 0
3 years ago
se the drawing tool to form the correct answers on the provided grid. For the sequence (1, 2, 4, 8, . . .), the index starts at
Andru [333]

Answer:

See attached

Step-by-step explanation:

For each increase of 1 in the index, the function value doubles. The size of the function value rapidly exceeds the limit of any linear grid.

We've shown a few points plotted. Perhaps it will give you the idea of what you need to do.

6 0
3 years ago
PLZ NEED HELP RIGHT AWAY
Sever21 [200]
1) A
3) B
4) A
5)A
sorry that i dont have all the answers, hope this helped
8 0
3 years ago
Read 2 more answers
Please help me Simplify: 6! − 3!
djverab [1.8K]

Answer: Option B 3!((6×5×4)-1)

Step-by-step explanation:

6! - 3!

= (1×2×3×4×5×6) - (1×2×3)

= 720 - 6

= 714

As you told you forgot to put options lemme answer it.

It will be option B

3!((6×5×4)-1)

It will be 6((6×5×4)-1)

= 6(120) -6

= 720 - 6

= 714

please click thanks and mark brainliest if you like :)

5 0
3 years ago
Read 2 more answers
If a = √3-√11 and b = 1 /a, then find a² - b²​
Serga [27]

If b=\frac1a, then by rationalizing the denominator we can rewrite

b = \dfrac1{\sqrt3-\sqrt{11}} \times \dfrac{\sqrt3+\sqrt{11}}{\sqrt3+\sqrt{11}} = \dfrac{\sqrt3+\sqrt{11}}{\left(\sqrt3\right)^2-\left(\sqrt{11}\right)^2} = -\dfrac{\sqrt3+\sqrt{11}}8

Now,

a^2 - b^2 = (a-b) (a+b)

and

a - b = \sqrt3 - \sqrt{11} + \dfrac{\sqrt3 + \sqrt{11}}8 = \dfrac{9\sqrt3 - 7\sqrt{11}}8

a + b = \sqrt3 - \sqrt{11} - \dfrac{\sqrt3 + \sqrt{11}}8 = \dfrac{7\sqrt3 - 9\sqrt{11}}8

\implies a^2 - b^2 = \dfrac{\left(9\sqrt3 - 7\sqrt{11}\right) \left(7\sqrt3 - 9\sqrt{11}\right)}{64} = \boxed{\dfrac{441 - 65\sqrt{33}}{32}}

5 0
2 years ago
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