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ra1l [238]
3 years ago
11

944-6552-9711 psw zV3WZ3 bo y ss

Mathematics
1 answer:
NemiM [27]3 years ago
3 0

Answer:

Uhm okay

Step-by-step explanation:

Wanna talk I’m bored :/

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Eliza fell asleep at 9:35 pm she awoke at 8 hours 45 minutes later what time did she wake up??
zzz [600]

Answer:

6:20 AM

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Can someone please help me I don’t know how to do this (Due today)
kirill115 [55]

First go to the y intercept (or the b in y=mx+b) look for the slope and plot the points on the graph they're talking about e.g. #23 the y-intercept is 6 go to the 6 on the y axis, and then look at the slope (x), so it goes up and over to the right since it's positive by 1

6 0
3 years ago
The description below represents Function A and the table represents Function B:
frosja888 [35]
Hello there!

It would be their slopes are equal but y-intercepts are not equal.


Hope This Helps You!
Good Luck :)
5 0
4 years ago
Bonnie is n years old.
grin007 [14]

Answer:

T = 5n + 7

Step-by-step explanation:

Bonnie : n

Clara : 3n

Danesh : 7 + n

T = n + 3n + 7 + n

T = 5n + 7

6 0
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

4 0
4 years ago
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