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cestrela7 [59]
3 years ago
6

Im h--> 0 ((Sqrt 9+h)-3)/h

Mathematics
1 answer:
Svetradugi [14.3K]3 years ago
6 0
\lim_{h \to 0}  \frac{ \sqrt{9+h}-3}{h} = \frac{ \sqrt{9+0} -3}{0}= \frac{0}{0} \\  \lim_{h \to 0}  \frac{ \sqrt{9+h}-3}{h}* \frac{ \sqrt{9+h}+3 }{ \sqrt{9+h} +3}= \\ = \lim_{h \to 0}   \frac{9+h-9}{h( \sqrt{9+h} +3)}  = \\   \lim_{n \to 0} \frac{h}{h( \sqrt{9+h}+3) }= \\ = \frac{1}{ \sqrt{9} +3}= 
= 1/6
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Solve for x ????????????
Novosadov [1.4K]

Answer:

x = 8°

Step-by-step explanation:

since the angles are vertical angles, they are equal

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Solve for x. round to the nearest tenth of a degree, if necessary.​
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How Many times do at least 2 odd numbers show up when you count from 0 to 999? Show your work.
postnew [5]

let's count from 0 to 999

0, 1, 2, 3, ooh, look, there are 2 odd numbers, 4, 5, look another one, 6, 7, another!


we can see that every time you count from 0 to 999, at least 2 odd numbers show up

so answer is 1 time  (because if 2 odd numbers or 20 odd numbers show up, it's still at least 2 odd numbers)




I think you really want to know how many times a pair off numbers occur together

like 33, 313 or 333, but not 31

that is interesting

in the first 10's 11 is only one

in each odd 10, in the 100's reigon, there is only 1 pair of odds

so 11, 33, 55, 66, 77, 99 or 6 pairs in the pre-100's zone

in the 100's zone, we have 101, 110-119, 121, 131, 141, 151, 161, 171, 181, 191 or 19 numbers with pairs of odds

in 200's, it's the same as pre-100's, 6 pairs

in the 300's zone, it's like 100's so 19 pairs

in the 400 zone, it's like 200's, 6 pairs

in 500's zone, it's like 100's, so 19 pairs

in the 600 zone, it's like 200's, 6 pairs

in 700's zone, it's like 100's, so 19 pairs

in the 800 zone, it's like 200's, 6 pairs

in 900's zone, it's like 100's, so 19 pairs

so a total of 6+19+6+19+6+19+6+19+6+19=125 pairs of odds

8 0
3 years ago
I need help please now!!!!
Nadusha1986 [10]

Answer:

top right one

Step-by-step explanation:

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The answer is in ur mom Virginia
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