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Viktor [21]
3 years ago
5

Which of the following points lie in the solution set to the following system of inequalities?

Mathematics
1 answer:
BartSMP [9]3 years ago
3 0
B it’s b it’s 1,5 I just took it it’s be
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12meters; is how many FT
kkurt [141]
12 meters is equal to 39.3701 feet.

1 meter is equal to 3.28084 feet.
4 0
3 years ago
In the 1970s, due to world events, there was a gasoline shortage in the United States. There were often long lines of cars waiti
Wittaler [7]

Answer:

A line of cars will be about 9650000 feet

Step-by-step explanation:

Simple:

9.65*1,000,000=9650000

Hope this Helps!

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6 0
3 years ago
A pack of cardinal flower seeds cost $4 and a pack of petunia flower seeds cost $2.50. you buy the same number of packs of each
jonny [76]
12 u have to divide i think
6 0
3 years ago
Kira unravels a shirt to get blue thread. From each centimeter (cm) of the shirt, Kira gets 231 cm of thread. She unravels 7cm o
cricket20 [7]

Answer:

1617 cm of thread.

Step-by-step explanation:

We are told that the equivalence would be that for each cm of shirt Kira gets 231 cm of thread, therefore it would be:

231 cm thread / 1 cm shirt

they tell us that she managed to remove 7 cm of shirt, therefore:

7 cm shirt * 231 cm thread / 1 cm shirt = 1617 cm thread

Therefore Kira obtained 1617 cm of thread.

6 0
2 years ago
We select n + 1 different integers from the set { 1 , 2 , ··· , 2 n } . Provethat there will alwaysbe two among the selected inte
just olya [345]

Answer:

See answer below

Step-by-step explanation:

From the set

{1,2,3,4...2n} we have 2n numbers in total , n are odd and n are even , therefore for a sample of n+1 numbers , we have at least 1 even number and 1 odd number.

Then

it the set includes 1 , the largest common divisor is 1 for 1 and the other numbers

if the set includes 3, there will be always a number that is not divisible by 3. Even we construct a set of n+1 numbers that are multiple of 3 , the largest number would be 3*(n+1)= 3*n+3 > 2*n (out of bounds) , therefore we are forced to take other number that is not divisible by 3  → the largest common divisor of that number with 3 is 1

If the set includes any other prime number → the largest common divisor of that with any other is 1

For the remaining odd numbers N, they can be factorised into other 2 odd common divisors N₂ and n₂ :

N = N₂*n₂ , since n₂ ≥ 2 →  N₂ < N

then the even N₂ also should be contained in the set

therefore also for N₂

N = N₃*n₃ →  N₃ < N₂

therefore if we continue , we would obtain a number  even Nn that has no smaller common divisors → since we cannot take all the multiples of N min ( because Nmin*(n+1)= Nmin*n+Nmin > 2*n for Nmin≥2) → there is at least a number in the sample of n+1 integers whose largest common divisor is 1

6 0
3 years ago
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