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2 years ago

SOMEONE ILL GIVE 100 POINTS

Mathematics

1 answer:

dsp732 years ago

4 0

Answer:

Step-by-step explanation:

1. ∠1 ≅∠3 1.Given

2. m∠1 = m∠3 1. Given

3. ∠1 and ∠2 are linear pair 3.) linear pair

4. ∠1 an ∠2 are supplementary 4. Definition of linear pair

5. ∠1 + ∠2 = 180 5. Definition of linear pair

6. ∠3 + ∠2 = 180 6.substitution property of equality

7. ∠2 an ∠3 are supplementary 7. Proven

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Determine what type of model best fits the given situation:

lyudmila [28]

Let value intially be = P

Then it is decreased by 20 %.

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)

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

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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