<span>The difference between two integers is at most 16
x1 - big integer
x2- small integer

</span><span>the smaller integer is 12
so
x2 = 12
substitution
</span>
<span>

solve for x1
however, note the fact larger
thus meaning
x1 > x2 applies to this problem
</span>
Answer:
Step-by-step explanation:
Hello
A (5;2)
B (-1;-7)
y = ax + b
2 = 5a + b
-7 = -a + b
2 - (-7) = 5a - (-a) + b - b
2 + 7 = 5a + a
9 = 6a
a = 9/6
a = 3/2
2 = 3/2 * 5 + b
b = 2 - 15/2
b = 4/2 - 15/2
b = 11/2
y = 3/2 x + 11/2
y+1= 3/2(x-4)
y + 1 = 3/2 x - 6
y = 3/2 x - 6 - 1
y = 3/2 x - 7 => no
y-4= 3/2(x+1)
y - 4 = 3/2 x + 3/2
y = 3/2 x + 3/2 + 4
y = 3/2 x + 3/2 + 8/2
y = 3/2 x + 11/2 => yes
y+4= 3/2(x-1)
y + 4 = 3/2 x - 3/2
y = 3/2 x - 3/2 - 4
y = 3/2 x - 3/2 - 8/2
y = 3/2 x - 11/2 => no
y-1= 3/2 (x+4)
y - 1 = 3/2 x + 6
y = 3/2 x + 6 + 1
y = 3/2 x + 7 => no
Answer: 1.5
Step-by-step explanation: The difference between the 2nd And 4th is 7. Divide by 2 to get 3.5. Add 3.5 to f(2) to getf(3)
The difference between f(1) and f(2) will be 3.5 and that's 5-3.5 =1.5
F(5) is 15.5 f(6) is 19
I believe they are about the similar in shape unless you have answers to them?
X=6
y=7
6+7
= 13
1/2(6) + 7
= 3 +7
=10