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gregori [183]
3 years ago
7

In the equation y = -2x + 5, what is the average rate of change (slope)?

Mathematics
2 answers:
Shalnov [3]3 years ago
8 0
Slope = -2

y= Mx+b M= slope
y=-2x+b. -2 is in the m spot
lorasvet [3.4K]3 years ago
3 0

Answer:

-2

Step-by-step explanation:

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3 years ago
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
agasfer [191]

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

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3 years ago
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<span>10x-3=6x+85
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6 0
3 years ago
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Step by step explanation
4 0
3 years ago
If a tank holds 5000 gallons of water, which drains from the bottom of the tank in 40 minutes, then Torricelli's Law gives the v
pochemuha

Answer:

V'(t) = -250(1 - \frac{1}{40}t)

If we know the time, we can plug in the value for "t" in the above derivative and find how much water drained for the given point of t.

Step-by-step explanation:

Given:

V = 5000(1 - \frac{1}{40}t )^2  , where 0≤t≤40.

Here we have to find the derivative with respect to "t"

We have to use the chain rule to find the derivative.

V'(t) = 2(5000)(1 - \frac{1}{40} t)d/dt (1 - \frac{1}{40}t )

V'(t) = 2(5000)(1 - \frac{1}{40} t)(-\frac{1}{40} )

When we simplify the above, we get

V'(t) = -250(1 - \frac{1}{40}t)

If we know the time, we can plug in the value for "t" and find how much water drained for the given point of t.

4 0
3 years ago
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