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zvonat [6]
2 years ago
12

3. What is the GCF of 12, 24, and 36? A. 12 C. 24 B. 18 D. 48​

Mathematics
2 answers:
agasfer [191]2 years ago
7 0

Answer:

A

Step-by-step explanation:

GCF is greatest common factor, which means that it is the greatest number all the numbers can be divisible by

for these numbers it is 12

Ludmilka [50]2 years ago
6 0

Hey there!

Finding the GCF (Greatest Common Factor) of 12, 24, & 48

List of the factors

• 12: 1, 2, 3, 4, 6, & 12

• 24: 1, 2, 3, 4, 6, 8, 12, & 24

• 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, & 48

Common factors

• 1, 2, 3, 4, 6, & 12

Biggest numbers out of the factors

• 6 & 12

GREATEST common factor

• 12

Therefore, your answer is: GCF (Greatest Common Factor) → 12

• Option A.

Good luck on your assignment and enjoy your day!

~Amphitrite1040:)

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Does there exist a di↵erentiable function g : [0, 1] R such that g'(x) = f(x) for all x 2 [0, 1]? Justify your answer
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Answer:

No; Because g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Step-by-step explanation:

Assuming:  the function is f(x)=x^{2} in [0,1]

And rewriting it for the sake of clarity:

Does there exist a differentiable function g : [0, 1] →R such that g'(x) = f(x) for all g(x)=x² ∈ [0, 1]? Justify your answer

1) A function is considered to be differentiable if, and only if  both derivatives (right and left ones) do exist and have the same value. In this case, for the Domain [0,1]:

g'(0)=g'(1)

2) Examining it, the Domain for this set is smaller than the Real Set, since it is [0,1]

The limit to the left

g(x)=x^{2}\\g'(x)=2x\\ g'(0)=2(0) \Rightarrow g'(0)=0

g(x)=x^{2}\\g'(x)=2x\\ g'(1)=2(1) \Rightarrow g'(1)=2

g'(x)=f(x) then g'(0)=f(0) and g'(1)=f(1)

3) Since g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Because this is the same as to calculate the limit from the left and right side, of g(x).

f'(c)=\lim_{x\rightarrow c}\left [\frac{f(b)-f(a)}{b-a} \right ]\\\\g'(0)=\lim_{x\rightarrow 0}\left [\frac{g(b)-g(a)}{b-a} \right ]\\\\g'(1)=\lim_{x\rightarrow 1}\left [\frac{g(b)-g(a)}{b-a} \right ]

This is what the Bilateral Theorem says:

\lim_{x\rightarrow c^{-}}f(x)=L\Leftrightarrow \lim_{x\rightarrow c^{+}}f(x)=L\:and\:\lim_{x\rightarrow c^{-}}f(x)=L

4 0
3 years ago
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