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stira [4]
3 years ago
15

E^x = 25 ...................

Mathematics
2 answers:
oksano4ka [1.4K]3 years ago
6 0
2^{x} =4 ----\ \textgreater \  x =  log_{2} 4
anzhelika [568]3 years ago
5 0
X is equal to about 3.218876
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that would be 38 degrees

Step-by-step explanation:

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The predicted calories in food items based on grams of fat are represented by the residual plot.
VashaNatasha [74]

Answer:

The correct option is;

C. The pattern is random, indicating a good fit for a linear model

Step-by-step explanation:

A graph that has the residuals (the difference between the value observed and the value expected (regression analysis) on the vertical axis and the variable that is not affected by the other variables (independent variable) on the x or horizontal axis is known as a residual plot

A linear regression model is suited in a situation where the points are dispersed randomly on both sides of the horizontal axis

Therefore, given that the first point is below the horizontal axis and the next point is above the horizontal axis, while the third and the fourth points are below the horizontal axis, the fifth, sixth, and seventh points are above the horizontal axis and the eighth point is below the horizontal axis, the points are random around the horizontal axis, indicating the suitability of a linear regression model.

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grin007 [14]

Answer:t the farthest city south is bellville and the volcano name is sue

Step-by-step explanation:

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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
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

4 0
4 years ago
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