24 ÷ b +8 when B is 9
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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
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m=18 when r = 2.
Step-by-step explanation:
Given,
m∝
So,
m = k×,--------eq 1, here k is the constant.
To find the value of m when r = 2
At first we need to find the value of k
Solution
Now,
Putting the values of m=9 and r = 4 in eq 1 we get,
9 =
or, k = 36
So, eq 1 can be written as m=
Now, we put r =2
m =
or, m= 18
Hence,
m=18 when r = 2.