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The true about the domain and the range of the function is:
The domain is all real numbers, and the range is all real numbers
greater than or equal to -4 ⇒ 1st answer
Step-by-step explanation:
f(x) = (x + 2)(x + 6) is a quadratic function with 2 factors (x + 2) and (x + 6)
By multiplying its two factors we will find the form of the quadratic function
∵ (x)(x) = x²
∵ (x)(6) = 6x
∵ (2)(x) = 2x
∵ (2)(6) = 12
∴ f(x) = x² + 6x + 2x + 12
- By adding like terms
∴ f(x) = x² + 8x + 12
The quadratic function represented graphically by a parabola
Look to the attached figure
The x-coordinate of the vertex point of the parabola h =
where b is the coefficient of x and a is the coefficient of x²
∵ f(x) = x² + 8x + 12
∴ a = 1 and b = 8
∴ h =
The y-coordinate of the vertex point is k = f(h)
∵ h = -4
∴ k = f(-4)
∴ k = (-4)² + 8(-4) = 12 = 16 - 32 + 12
∴ k = -4
∴ The vertex point of the parabola is (-4 , -4)
∵ The parabola is opened upward
∴ Its vertex is minimum point
∴ The minimum value of f(x) is y = -4
∵ The domain of the function is the values of x
∵ The range of the function is the values of y corresponding to the
values of x
∵ x can be any real numbers
∴ x ∈ R, where R is the set of real numbers
∴ The domain of f(x) is all real numbers
∵ The minimum value of f(x) is y = -4
∴ y can be any real number greater than or equal to -4
∴ y ≥ -4
∴ The range is all real number greater than or equal to -4
The true about the domain and the range of the function is:
The domain is all real numbers, and the range is all real numbers
greater than or equal to -4
Learn more:
You can learn more about quadratic function in brainly.com/question/1332667
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This question is incomplete, the complete question is;
If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple are written in increasing order but are not necessarily distinct.
In other words, how many 5-tuples of integers ( h, i , j , m ), are there with n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1 ?
Answer:
the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120
Step-by-step explanation:
Given the data in the question;
Any quintuple ( h, i , j , m ), with n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1
this can be represented as a string of ( n-1 ) vertical bars and 5 crosses.
So the positions of the crosses will indicate which 5 integers from 1 to n are indicated in the n-tuple'
Hence, the number of such quintuple is the same as the number of strings of ( n-1 ) vertical bars and 5 crosses such as;
= [( n + 4 )! ] / [ 5!( n + 4 - 5 )! ]
= [( n + 4 )!] / [ 5!( n-1 )! ]
= [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120
Therefore, the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120