4 divided by 457. Please list the steps

2 answers:

5 0

You find out how many times 4 goes into 4 and get 1 then you find out how many times 4 goes into 5 and get 1 but you subtract 5 minus 4 and get 1 then bring down the 7 to get 17 then find out how many times 4 goes into 17 which is 4 times because 4 times 4 is 16 and you do 17 minus 16 and get 1 then add a decimal and bring down the 1 to get 10 then find out how many times 4 goes into 10 and get 2 subtract and get 2 bring down the zero turn it into 0 then you get 20 then find out how many times 4 goes into 20 and get 5 and 4*5=20 so 20-20 is 0 so your answer is 114.25

wel3 years ago

5 0

first you start your thousands place
4÷4
do that then you'll end up with 0
then, bring down the 5
put 1 at the top
4×1=4
5-4=1
bring down the 7
1 1 4
|--------
4| 457
- 4⬇|
-----------|
0 5 |
- 4 |
------------|
17
- 16
--------------
1
114 r1

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2/3* (1/3+3 1/4 help pls now

ale4655 [162]

Answer:

43/18

Step-by-step explanation:

2/3(1/3+ 3 1/4)

First solve parenthesis.

2/3(1/3+ 3 1/4)

find the lowest common denominator (12) to add the two fractions.

1/3= 4/12;

3 1/4= 13/4 = 39/12

4/12+ 39/12= 43/12

2/3(43/12)

Then multiply the fractions.

2/3(43/12)

Numerator: 2*43= 86

Denominator: 3*12= 36

86/36

Last simplify your fraction.

43/18 or 2 7/18

4 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).

$f'(c)=\lim_{x\rightarrow c}\left [\frac{f(b)-f(a)}{b-a} \right ]\\\\g'(0)=\lim_{x\rightarrow 0}\left [\frac{g(b)-g(a)}{b-a} \right ]\\\\g'(1)=\lim_{x\rightarrow 1}\left [\frac{g(b)-g(a)}{b-a} \right ]$