2. Given two consecutive positive EVEN integers, the square of the largest minus
10 times the smaller is 76. What is the sum of the two integers?

26

Step-by-step explanation:

2. Given two consecutive positive EVEN integers, the square of the largest minus 10 times the smaller is 76. What is the sum of the two integers?

I'll call the two integers a and b.    a=b+2

a² - 10b = 76

(b+2)² - 10b =76

(b+2)(b+2) -10b = 76

b² +2b + 2b +4  -10b = 76

b² +4b + 4  -10b  = 76

b² -6b + 4 =76  subtract 76 from both sides to get the right side equal to zero

b² -6b +4 -76 = 0

b² -6b - 72 = 0  what factors of -72 add up to -6? The two that work are -12, 6

(b-12)(b+6) = 0   so  b-12  = 0   or b+6  =  0

                                 +12  +12         -6     -6

                                  b = 12      or     b = -6  but b must be positive so b=12, a=14  

a+b= 14+12 = 26

38.5

step-by-step explanation:

lets say, first number = r

second number = 3r+7

total = 49

(r)+(3r+7)= 49

4r+7 = 49

4r = 42

r = 42/4

r= 10.5

3r+7 = 3(10.5)+7 = 38.5

hope this

arithmetic sequence

step-by-step explanation:

given that a sequence has terms as

19, 15, 11,

this is represented on a graph

the pattern indicates that the sequence is an arithmetic sequence with i term

a=19 and common difference d =-4

any general term

thus the sequence is reduced by 4 for each successive term starting from 1 and nth term is

24-5n.

first negative term would start after 5th term

5th term =24-25< 0

from 5th term onwards the sequence would be negative.