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Dafna11 [192]
2 years ago
11

Write the ratio of 14 and 18 in its lowest terms

Mathematics
2 answers:
olga_2 [115]2 years ago
7 0

Answer:

the correct answer is 7 and 9

ivolga24 [154]2 years ago
3 0

Answer:

7 and 9

Step-by-step explanation:

14 ÷ 2 and 18÷ 2

= 7 and 9

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PLEASE HELP ME IM SO CONFUSED!!! i need answers 10-12
zimovet [89]

Answer: Heyaa! ~

   10. 8√2

   11. 2h²√3

   12. 2h³√5k⁴

Step-by-step explanation:

    <em>- Lets solve it together! </em>

Simplify the radical by breaking the radicand up into a product of known factors, assuming positive real numbers.

Hopefully this helps you!

3 0
2 years ago
How many kilometers?
Anettt [7]

Answer:

7.12 kilometers

Step-by-step explanation:

Add the amounts of kilometers to find the total :

3.8 kilometers + 3.32 kilometers = 7.12 kilometers

8 0
3 years ago
A cook mixes 4 pounds of rice into 5 quarts of boiling water.
professor190 [17]
I think you could times 4 and 5 to get 20
5 0
3 years ago
Read 2 more answers
At the local college, a study found that students earned an average of 14 credit hours per semester. A sample of 138 students wa
77julia77 [94]

Answer:

1932 credit hours

Step-by-step explanation:

Create a proportion

credits             credits

----------     =      ----------

students          students

14      ?

---- = -----

1         138

14 times 138

1932 credit hours

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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