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2 years ago

Any one help me with this plsss and ni links plss!!

Mathematics

1 answer:

Bas_tet [7]2 years ago

4 0

Answer:

6. D.

7. F.

8. A.

9. B.

10. C.

Step-by-step explanation:

6. 9 + (12 - 10)

12 - 10 = 2

9 + 2 = 11

7. (20 - 15) x 2

20 - 15 = 5

5 x 2 = 10

8. 10 ÷ 5 + 7

10 ÷ 5 = 2

2 + 7 = 9

9. 6 + 2 x 3

2 x 3 = 6

6 + 6 = 12

10. (2 x 4) + 8

2 x 4 = 8

8 + 8 = 16

Let me know if this helps!

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Determine what type of model best fits the given situation:

lyudmila [28]

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So 20% of P = $\frac{20}{100} \times P = 0.2P$

So after 1 year value is decreased by 0.2P

so value after 1 year will be = P - 0.2P (as its decreased so we will subtract 0.2P from original value P) = 0.8P-------------------------------------(1)

Similarly for 2nd year, this value 0.8P will again be decreased by 20 %

so 20% of 0.8P = $\frac{20}{100} \times 0.8P = (0.2)(0.8P)$

So after 2 years value is decreased by (0.2)(0.8P)

so value after 2 years will be = 0.8P - 0.2(0.8P)

taking 0.8P common out we get 0.8P(1-0.2)

= 0.8P(0.8)

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Similarly after 3 years, this value $P(0.8)^{2}$ will again be decreased by 20 %

so 20% of $P(0.8)^{2} \frac{20}{100} \times P(0.8)^{2} = (0.2)P(0.8)^{2}$

So after 3 years value is decreased by $(0.2)P(0.8)^{2}$

so value after 3 years will be = $P(0.8)^{2} - (0.2)P(0.8)^{2}$

taking $P(0.8)^{2}$ common out we get $P(0.8)^{2}(1-0.2)$

$P(0.8)^{2}(0.8)$

$P(0.8)^{3}$ -----------------------(3)

so from (1), (2), (3) we can see the following pattern

value after 1 year is P(0.8) or $P(0.8)^{1}$

value after 2 years is $P(0.8)^{2}$

value after 3 years is $P(0.8)^{3}$

so value after x years will be $P(0.8)^{x}$ ( whatever is the year, that is raised to power on 0.8)

So function is best described by exponential model

$y = P(0.8)^{x}$ where y is the value after x years

so thats the final answer

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