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ankoles [38]
3 years ago
10

I don’t know the answer

Mathematics
1 answer:
Andreas93 [3]3 years ago
5 0

I believe the answer would be yes. Since your dealing with Ratios it would be simple to understand that they are proportional. I'll explain a little better.

This means that if you double this ratio, it will still be equivalent to 1:3

1:3

This is also proportional because this number will always be equivalent to itself.

7:21

-----------------------------------------------------------------------------------------------------------------Hope this helps

-Miri

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It takes you 45 minutes to clean the fish tank and it takes your sister 30 minutes. how long does it take the two of you working
katen-ka-za [31]

It takes you 45 minutes to clean the fish tank and it takes your sister 30 minutes. how long does it take the two of you working together to clean the fish tank

Answer:

My 1 minute work =\frac{1}{45}

My sisters 1 minute work =\frac{1}{30}

My and My sister's 1 minute work =\frac{1}{45}+\frac{1}{30}

                                                        =\frac{5}{90}

                                                        =\frac{1}{18}

\therefore Me and My sister will finish the work in \frac{18}{1} =18 minutes

5 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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lorasvet [3.4K]

Answer:

No.

Step-by-step explanation:

The Triangle Inequality Theorem states that this is not possible.

8 0
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12 2/5 x 3 1/6 pls help me
Maurinko [17]
1. Turn improper
12 \frac{2}{5}  =  \frac{62}{5}
3 \frac{1}{6}  =  \frac{19}{6}
2. set up the problem
\frac{62}{5}  \times  \frac{19}{6}
now multiple across and simplefy
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3 years ago
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Answer: 13/8 or 1.625

Step-by-step explanation:

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