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Sidana [21]
3 years ago
12

Find out 6 rational numbers between 3 and 4

Mathematics
2 answers:
Alisiya [41]3 years ago
7 0
13/4,14/4,15/4,19/6,20/6,21,6
mixer [17]3 years ago
6 0

Answer:

22/7, 23/7, 24/7, 25/7, 26/7, and 27/7.

Step-by-step explanation:

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Write an expression for 5 times the difference of twice a number and 4
777dan777 [17]

Answer:

1. 50-x

2. 2x+189

3. x+109

4. x+99

5. x*(76-39)

6. 75+x

7. 45-3x

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Step-by-step explanation:

4 0
2 years ago
Please solve the table and show your work
Elanso [62]

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a

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3 years ago
Select the two values of x that are roots of this equation.<br> 2x-5=-3x2
anygoal [31]

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Step-by-step explanation:

4 0
2 years ago
What is the distance between (-5, 3) and (7, 3)?
Rama09 [41]

Answer:

12 units

Step-by-step explanation:

We would need to ignore the 3 first of all. We know that 3 is located at the same place. We then need to see 7 and -5. The distance from zero of each numbers matter, not the ammount. So (-5) is 5 from zero. (7) is 7 from zero. 5 + 7 = 12 units

4 0
3 years ago
If limx→3f(x)=7, which of the following must be true? I. f is continuous at x = 3 II. f is differentiable at x = 3
Viktor [21]

Without knowing anything else about f(x), neither of these need be true.

Suppose

f(x)=|x-3|+7=\begin{cases}10-x&\text{for }x3\\0&\text{for }x=3\end{cases}

Then (I) isn't true because, while the limit exists as x\to3 and is equal to 7, we have f(3)=0\neq7, so f(x) is not continuous there.

(II) also true because f(x) is not differentiable at x=3; that is,

\displaystyle\lim_{x\to3^-}f'(x)=\lim_{x\to3}(10-x)'=\lim_{x\to3}(-1)=-1

but

\displaystyle\lim_{x\to3^+}f'(x)=\lim_{x\to3}(x+4)'=\lim_{x\to3}1=1

which means the derivative does not exist at x=3.

3 0
3 years ago
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