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AlexFokin [52]
4 years ago
5

Find three consecutive odd integers whose sum is 13 more than twice the largest of the three intergers

Mathematics
1 answer:
Yuki888 [10]4 years ago
4 0

15,17,19 
Suppose the three odd integers are #n#, #n+2# and #n+4# 
Their sum is: 
#n + (n+2) + (n+4) = 3n+6# 
#13# more than twice the largest of the three is: 
#2(n+4)+13 = 2n+21# 
From what we are told these two are equal: 
#3n+6 = 2n+21# 
Subtract #2n+6# from both sides to get: 
#n = 15# 
So the three integers are: 
#15, 17, 19#
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What is the common difference of 31, -169, -369, -569?
san4es73 [151]

Answer:

31 is a positive number, while -169, -369, and -569 is negative.

Step-by-step explanation:

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Positive numbers are numbers above 0 and don't include a negative sign.

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4 0
3 years ago
Help me out please T-T​
Anettt [7]

Answer:

The simplyfied version would be 19/4

Show of work:

(1/4)^-2 = 4^2

3 × 8^2/3 × 1 = 12

(9/16)^1/2 = 3/4

4^2 - 12 + 3/4

Convert elements to fractions:

-12 × 4 + 3

---------- ----

4 4

Since the denominators are equal combine the fractions:

-12 × 4 + 3

---------------

4

-12 × 4 + 3 = -45

= -45/4

=4^2 - 45/4

4^2 = 16

16 - 45/4

16 × 4 - 45. 16 × 4 - 45

--------- ----- ----------------

4 4. 4

-> 16 × 4 - 45 = 19

= 19/4

7 0
3 years ago
Read 2 more answers
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