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kirza4 [7]
3 years ago
5

What 6.1 divided by eight

Mathematics
2 answers:
geniusboy [140]3 years ago
4 0
6.1 divided by 8 = 0.7625
8_murik_8 [283]3 years ago
3 0
The answer is 0.7625
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What is the prime factorization for the number 20 using exponents
Alexxandr [17]
Prime factorization of 2·5·2=2^2·5
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3 years ago
in a regular polygon each interior angle is 140 greater than each exterior angle calculate the number of sides of each polygon .
Elodia [21]

Answer:

9

Step-by-step explanation:

Exterior sides =  180 - 140 = 40

number of sides = 360 / 40 = 9 sides

6 0
3 years ago
What does 10+q equal?
BlackZzzverrR [31]

Answer:

Well if no variable substitution is given its 10+q

10+q=10+q

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2 years ago
Read 2 more answers
Write an expression that is equivalent to 3/5 (3y+15)<br> 3/5y +9, 3/5y+15, 9/5y+9, 9/5y + 15?
NISA [10]
Given the expression

\frac{3}{5} (3y+15)

The expression that is equivalent to the given expression is given by:

\frac{3}{5} (3y+15)= \frac{9}{5} y+ \frac{3}{5} (15)= \frac{9}{5} y+9
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
3 years ago
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