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Natasha_Volkova [10]
1 year ago
8

Given the function g(x)=3x-7, determine when g(x)=-4

Mathematics
1 answer:
sp2606 [1]1 year ago
3 0

You have the following function:

g(x) = 3x - 7

In order to determine th value of x when g(x) = -4, you equal the previous expression to -4 and solve for x, just as follow:

g(x) = 3x - 7 replace g(x) = -4

-4 = 3x - 7 add 7 both sides

-4 + 7 = 3x simplify

3 = 3x divide by 3 both sides

3/3 = x

x = 1

Hence, the anser is x = 1

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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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3 years ago
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Answer:

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Step-by-step explanation:

<u>Inequality "y is less than or equal to 5 and is greater than or equal to -2":</u>

  • - 2 ≤ y ≤ 5

<u>This is the interval:</u>

  • y ∈ [- 2, 5]

<u>The endpoints are:</u>

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<u>Product of the endpoints:</u>

  • - 2 × 5 = - 10

<u>Answer choices:</u>

  • (A) odd ⇒ False, - 10 is even
  • (B) positive ⇒ False, - 10 is not positive
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  • (D) None of the above​ ⇒ False, option C is correct
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Step-by-step explanation:

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Step-by-step explanation:

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