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Inga [223]
2 years ago
15

Pls help will give brainliest

Mathematics
1 answer:
sveticcg [70]2 years ago
4 0

3.06 i believe is the answer! im not sure but i've calculated

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Order the following numbers from least to greatest. √11, -0.4, (-4/3), 0.8, √2
agasfer [191]

Answer:

(-4/3), 0.4, 0.8, √2, √11

Step-by-step explanation:

√11=3.316624790355399849114932736670686683927088545589353597058

0.4

(-4/3) = -1.33333333333333333333333333333333333333333333333333333333

0.8

√2=1.414213562373095048801688724209698078569671875376948073176


Thus :

(-4/3)

0.4

0.8

√2

√11

8 0
3 years ago
Please help will give brainlyist
Shalnov [3]
Number 11
Since the auditorium can hold 600 and the student body will be divide by four parts
The answer is
x ≤ 2400
6 0
3 years ago
Before a bag of flour can be sold, it must be 40 kilograms but can be within 0.5
Serjik [45]

Answer:

The minimum acceptable weight of the flour bag is 39.5 kg,

and its maximum acceptable weight is 40.5 kg

Step-by-step explanation:

Notice that the selling weight of the bag to be sold (x) needs to differ from 40 kg at most 0.5 kg. Therefore, we can write the following inequality:

|x-40|\leq 0.5

The inequality can be solve once we remove the absolute value symbol, and solve for the unknown "x". Recall that in order to solve it, we need to consider the two possible cases:

a) that the expression within the absolute value symbol is larger than or equal to zero, and b) that the expression is less than zero.

a) If x-40\geq 0 then its absolute value is equal to itself (x-40), and when we remove the absolute value symbols, we get:

x-40\leq 0.5\\x\leq 0.5+40\\x\leq 40.5\,\,kg

which means that the weigh "x" of the flour bag should be smaller or equal than 40.5  kg

b) If x-40, then the absolute value of this is its opposite : -x+40, and the inequality becomes:

-x+40\leq 0.5\\40\leq 0.5+x\\40-0.5\leq x\\39.5\leq x

which means that the weight of the flour bag must be larger than or equal to 39.5 kg

So, the minimum acceptable weight of the flour bag is 39.5 kg, and the maximum acceptable weight is 40.5 kg

6 0
3 years ago
If there are 7 basketballs for every 3 students at Tuckahoe, then how many students are there if the store room has 49 basketbal
mixas84 [53]

Answer:

21 Students

Step-by-step explanation:

7:3

7*7=49

3*7=21

49:21

6 0
3 years ago
Read 2 more answers
this is integrated math2, i’m so confused what does it mean and what is the answer help please (this is the last one lol sorry)
swat32
3x +2x+ 1+ x+ 5= 180
6 0
2 years ago
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