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Arada [10]
1 year ago
13

Graph the following three equations on your own device and answer the questions below.

Mathematics
1 answer:
Alina [70]1 year ago
5 0

Due to length restrictions, we kindly invite to read the explanation of this question for further details on functions.

<h3>What are the characteristics of each of the three functions?</h3>

a) In this part we must evaluate each of the three functions at the given value of x:

Case 1

f(10) = 5 · 10 + 7

f(10) = 57

Case 2

f(10) = 10² + 6

f(10) = 106

Case 3

f(10) = 2¹⁰ + 3

f(10) = 1027

b) In this part we must evaluate each of the three functions at the given value of x:

Case 1

f(100) = 5 · 100 + 7

f(100) = 507

Case 2

f(100) = 100² + 6

f(100) = 10006

Case 3

f(100) = 2¹⁰⁰ + 3

f(100) = 1.267 × 10³⁰ + 3

c) In this part we must evaluate each of the three functions at the given value of x:

Case 1

f(1000) = 5 · 1000 + 7

f(1000) = 5007

Case 2

f(1000) = 1000² + 6

f(1000) = 1000006

Case 3

f(1000) = 2¹⁰⁰⁰ + 3

f(1000) = (1.268 × 10³⁰)¹⁰ + 3

f(1000) = 10.744 × 10³⁰⁰ + 3

f(1000) = 1.074 × 10³⁰¹ + 3

e) The <em>third</em> function increases the fastest.

f) In this part we need to compare the <em>third</em> function with respect to the <em>first</em> and <em>second</em> functions:

5 · x + 7 = 2ˣ + 3

2ˣ - 5 · x   = 4

The solutions of the equation are x = - 0.675 and x = 4.81. The function will exceed the other <em>first</em> function at x = 4.81.

x² + 6 = 2ˣ + 3

2ˣ - x² = 3

The solution of the equation is x = 4.588. The function will exceed the other <em>second</em> function at x = 4.588.

g) Yes, <em>exponential</em> functions with bases greater than 1 will surpass <em>polynomic</em> function at any point x such that x > 0.

h) The domain represents the set of x-values of a function and the range represents the set of y-values of a function. Then, the domain and range of each function is:

Case 1

Domain - All <em>real</em> numbers.

Range - All <em>real</em> numbers

Case 2

Domain - All <em>real</em> numbers.

Range - [6, +∞)

Case 3

Domain - All <em>real</em> numbers.

Range - (3, +∞)

To learn more on functions: brainly.com/question/12431044

#SPJ1

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Two numbers have a sum of 1,212 and a difference of 518. What are the two numbers?
guapka [62]

Answer:

The two numbers are

518

694

Step-by-step explanation:

As per requirements the two equations will be

X+Y=1212

X-Y=518

When we subtract these equations we get

X=1212-518

X=694

Now put the X value and find the Y value in any equation

X+Y= 1212

Y=1212-X

Where X = 694

Y=1212-694

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X=694 and Y=518

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3 years ago
What is 9 2/5-8 1/3 I need to know for math tomorrow
patriot [66]

Answer:

1\frac{1}{15}

Step-by-step explanation:

Let's use a start by using common denominators. 9\frac{2}{5} = 9\frac{6}{15} and 8\frac{1}{3} = 8 \frac{5}{15}

Now, replace the original equation with the common denominators. 9\frac{6}{15} - 8\frac{5}{15}

Solve! 9\frac{6}{15} - 8\frac{5}{15} = 1\frac{1}{15}

8 0
3 years ago
Which expression shows how to convert 60 hours to days​
torisob [31]

Answer

60 hours = 2.5 days

Step-by-step explanation:

1 day = 24 hours

2 days = 48 hours

3 days = 72 hours

so 60 hours in = to 2.5 days

hope this helps

7 0
3 years ago
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Need help asap
harina [27]

Answer:

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Step-by-step explanation:

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2 years ago
A set of composite numbers less than 12 .express it in listing and set builder methods​
marusya05 [52]

Given:

A set of composite numbers less than 12.

To find:

The set by listing and set builder methods​.

Solution:

Composite number: A positive integer is called a composite number if it has at least one divisor other than 1 and itself.

Composite numbers less than 12 are 4, 6, 8, 9, 10.

By using the listing method , the given set can be expression as:

Set = {4, 6, 8, 9, 10}

The numbers 4, 6, 8, 9, 10 are greater than 1, less than 12 and non prime numbers.

By using set builder method, the given set can be expression as:

Set = \{x|x\neq P,1

Therefore, the list and set builder notation of the given set are {4, 6, 8, 9, 10} and \{x|x\neq P,1 respectively.

5 0
2 years ago
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