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liraira [26]
3 years ago
10

Determine the two closest integers for each irrational number also 117 not in pic

Mathematics
1 answer:
Grace [21]3 years ago
6 0
1.) 4.358898944
2.) 7.071067812
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Give two ways to write an expression as a phrase 6+ P
dangina [55]

1.   P+(3x2)

2.  P+3+3

I hope this helps!

5 0
3 years ago
find the consecutive even integers such that three times the larger number is 30 more than the smaller one. help :')
Minchanka [31]

Answer:

12 and 14

Step-by-step explanation:

Let the even integers be x and x+2

three times the larger number is expressed as 3(x+2)

30 more than the smaller one is x + 30

Equating both expressions

3(x+2) = x +30

Find x

3x+6 = x+30

3x-x = 30-6

2x = 24

x = 12

The second integer is 12+2 = 14

Hence the required integers are 12 and 14

4 0
3 years ago
If f(x) = x2, and<br> g(x) = x – 1, then<br> f(g(x)) = [? ]x2+[?]+ [?]
Basile [38]

Answer:

f(g(x)) = x² - 2x + 1

Step-by-step explanation:

To find f(g(x)), substitute x = g(x) into f(x) , that is

f(g(x))

= f(x - 1)

= (x - 1)² ← expand using FOIL

= x² - 2x + 1

5 0
3 years ago
Find the missing number.<br><br> n − 9.01 = 3.86<br><br> n = ?
aleksandr82 [10.1K]
I think the missing number should be
12.87
3 0
4 years ago
Read 2 more answers
What is a possible solution for x+6≥10
ira [324]

Answer:

4,5,6

Step-by-step explanation:

*The sign of ≥ means that it must be greater or equal to 10 in this case

1). 3 isn't a possible solution due to 9 being less than 10

3+6_10

9_10

9<10

2). 4 is a possible solution due to the sum of 4 and 6 being equal to 10

4+6_10

10_10

10=10

3). 5 is a possible solution due to the sum of 5 and 6 being larger than 10

5+6_10

11_10

11>10

4).6 is a possible solution due to the sum of 6 and 6 being larger than 10

6+6_10

12_10

12>10

Hope this helped! (:

6 0
3 years ago
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