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Artyom0805 [142]
3 years ago
10

What are the domain and range of each relation? Drag the answer into the box to match each relation. A mapping diagram. Element

x contains negative 2, negative 1, 0, and 2. Element y contains negative 4, negative 2, and 2. Negative 2 maps to negative 4. Negative 1 maps to negative 2 and 2. Zero maps to negative 4. Two maps to 2. x y −4 −2 −2 1 1 4 4 4 Domain: {−4, −2, 1, 4} Range: {−2, 1, 4} Domain: {−2, −1, 0, 2} Range: {−4, −2, 2} Domain: {−4, −2, −1} Range: {−2, 0, 1}

Mathematics
1 answer:
ra1l [238]3 years ago
7 0

Answer:

Domain: {−2, −1, 0, 2} Range: {−4, −2, 2}

Step-by-step explanation:

We have been given a mapping diagram as shown below:

X Y

-2 -4

-1 -2

0 2

2


For better view, you can check the attached mapping diagram.

From that diagram we have to find domain and range.

Domain contains only x values so domain will be:

Domain: {-2,-1,0,2}

Range contains only y values so range will be:

Range: {-4,-2,2}

We see that only third choice matches obtained values of domain and range.

Hence final answer is  Domain: {−2, −1, 0, 2} Range: {−4, −2, 2}

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tamaranim1 [39]

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b) t = 31.15

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  • A(t) is the amount of money after t years.  
  • P is the principal(the initial sum of money).  
  • r is the interest rate(as a decimal value).  
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In this problem:

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  • No information about the number of compounding per year, hence n = 1.

Item a:

P = 2000, hence:

A(t) = P\left(1 + \frac{r}{n}\right)^{nt}

A(8) = 2000\left(1 + \frac{0.0225}{1}\right)^{8}

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Item b:

A(t) = 4000, hence:

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A(3) = 2000, hence:

A(t) = P\left(1 + \frac{r}{n}\right)^{nt}

2000 = P\left(1 + \frac{0.0225}{1}\right)^{3}

P = \frac{2000}{(1.0225)^3}

P = 1870.85

A similar problem is given at brainly.com/question/24850750

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