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Ratling [72]
3 years ago
11

What is the distance between the origin and (−3,−4)? A. 5 B. 7 C. 12 D. 25

Mathematics
2 answers:
Alex17521 [72]3 years ago
5 0

The distance between the origin and (-3,-4) can be modeled with a triangle. The horizontal distance and vertical distance are two legs, and the distance between the origin and the point is the hypotenuse.

Since we are dealing with distance, we will use absolute values for the distances.

The origin is at (0,0), and the point is at (-3,-4). Add the x-values and y-values:

|0 + -3| = |-3| = 3

|0 + -4| = |-4| = 4

The two legs of the triangles are 3 and 4.

We'll use the Pythagorean theorem to find the distance. Square the two legs' lengths:

(3^2 + 4^2) = x^2 = (9 + 16) = 25 = x^2

Square root both sides to get x by itself:

\boxed{x = 5}

The distance between the origin and (-3,-4) will be 5.

lions [1.4K]3 years ago
4 0

To find the answer to this question you create a traingle using the origin and the plot given. Then when that is completed you count up which is  a. 5

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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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