just g o o g l e it. i'm sure g o o g l e knows ... ... ... ....
He <span>wants to write the fraction 4 / 6 as a sum of unit fractions. It is written as follows:
</span>
1/6 +1/6 +1/6+ 1/6= 4/6.
A unit fraction has a numerator of 1 so since 4 is the numerator in 4/6 you would have to add 1/6 for four times.
Hope this answers the question. Have a nice day.
F(x)=2x
G(x)=x+5
Now
F(g(x))=2(x+5)
=2x+10
G(f(x))=2x+5
Given the domain {-4, 0, 5}, what is the range for the relation 12x 6y = 24? a. {2, 4, 9} b. {-4, 4, 14} c. {12, 4, -6} d. {-12,
xz_007 [3.2K]
The domain of the function 12x + 6y = 24 exists {-4, 0, 5}, then the range of the function exists {12, 4, -6}.
<h3>How to determine the range of a function?</h3>
Given: 12x + 6y = 24
Here x stands for the input and y stands for the output
Replacing y with f(x)
12x + 6f(x) = 24
6f(x) = 24 - 12x
f(x) = (24 - 12x)/6
Domain = {-4, 0, 5}
Put the elements of the domain, one by one, to estimate the range
f(-4) = (24 - 12((-4))/6
= (72)/6 = 12
f(0) = (24 - 12(0)/6
= (24)/6 = 4
f(5) = (24 - 12(5)/6
= (-36)/6 = -6
The range exists {12, 4, -6}
Therefore, the correct answer is option c. {12, 4, -6}.
To learn more about Range, Domain and functions refer to:
brainly.com/question/1942755
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Answer:
(a) yes
(b) no; see below
Step-by-step explanation:
(a) Integer roots of the quartic will be integer divisors of 6. One of the divisors of 6 is 3, so 3 is a possible root.
(b) In order for 3 to be a double root, it would have to be a double factor of 6. The only integer factors of 6 are 1, 2, 3, 6. (3² = 9 is not one.)
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The quartic can be written as ...
k(x -a)(x -b)(x -c)(x -d) . . . . . where a, b, c, d, k are integers
The constant term will be kabcd, of which each of the roots is a factor. If the constant is 6 and one root is d=3, then we must have
kabcd = 3kabc = 6
kabc = 6/3 = 2
Among these four integer factors, there must be an even number of minus signs, and one that has the value ±2. Another root whose value is 3 will not satisfy the requirements.