B)4.3........................................
The equation of the line is:
y-yo = m (x-xo)
Where,
m = (y2-y1) / (x2-x1)
Substituting values:
m = (8-4) / (6-2)
m = (4) / (4)
m = 1
Then, the equation is:
y-4 = 1 * (x-2)
Rewriting:
y = x-2 + 4
y = x + 2
The midpoint is:
k = ((x1 + x2) / 2, (y1 + y2) / 2)
k = ((2 + 6) / 2, (4 + 8) / 2)
k = ((8) / 2, (12) / 2)
k = (4, 6)
Then, the equation of the perpendicular line that passes through k is:
y = -x + b
Looking for b we have:
6 = -4 + b
b = 6 + 4
b = 10
Substituting:
y = -x + 10
Answer:
An equation of a line perpendicular to jl and passing through k is:
y = -x + 10
The numbers that would be smaller would be the two with the difference of 26.
27-1=26 where as, if you were to use a 1 in each, 99+1= 100. Therefore 27 and 1 (28) is less than 99 and 1 (100).
the fiest answer is x=6 for the second one the answer is x=4
Step-by-step explanation:
x+15/5x+1 =15/35 then cross multiply
1st one
2a + 3b = 6 --> Equation 1
5a + 2b = 4 --> Equation 2
Equation 1 * 5 --> 10a + 15b = 30
Equation 2 * 2 --> 10a + 4b = 8
Equation 1 - Equation 2 --> 11b = 22 --> b = 2
Sub b=2 in Equation 1 (or 2, either works):
2a + 3b = 6
2a + 3(2) = 6
2a + 6 = 6
2a = 0
a = 0
b = 2, a = 0
