perpendicular lines have a slope that is a negative reciprocal
A) 4x-5y=5
subtract 4x
solve for y
-5y = -4x+5
divide by -5
y = 4/5 x+5 slope is 4/5 perpendicular slope is -5/4
y -y1 =m(x-x1) point slope form of a line
y-3 = -5/4 (x-5)
B) 5x+4y = 37
subtract 5x
4y =-5x +37
divide by 4
y =-5/4 x +37/4 slope is -5/4 perpendicular slope is 4/5
y -y1 =m(x-x1) point slope form of a line
y-3 = 4/5 (x-5)
C)4x+5y=5
subtract 4x
5y = -4x +5
divide by 5
y = -4/5 x +1
y =-4/5 x +1 slope is -4/5 perpendicular slope is 5/4
y -y1 =m(x-x1) point slope form of a line
y-3 = 5/4 (x-5)
D)5x-4y=8
subtract 5x
-4y = -5x+8
divide by -4
y = 5/4 x-2
y =5/4 x +-2 slope is 5/4 perpendicular slope is -4/5
y -y1 =m(x-x1) point slope form of a line
y-3 = -4/5 (x-5)
The given circles are given in standard form:
(x - xc)² + (y - yc)² = r²
The second quadrant is the one that has negative x coordinates and positive y coordinates.
This said, let's see all your options:
A) (x - 5)² + (y - 6)² = 25
xc = -(-5) = +5
yc = -(-6) = +6
C (5 , 6) is in the first quadrant.
B) (x + 1)² + (y - 7)² = 16
xc = -(+1) = -1
yc = -(-7) = +7
C (-1 , 7) is in the second quadrant.
C) (x - 4)² + (y + 3)² = 32
xc = -(-4) = +4
yc = -(+3) = -3
C (4, -3) is in the fourth quadrant.
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D) (x + 2)² + (y - 5)²= 9</span>
xc = -(+2) = -2
yc = -(-5) = +5
C (-2 , +5) is in the second quadrant.
Therefore, the correct answers are B and D.
Answer:
She did not use the reciprocal of the divisor.
She added the numerators.
The sup is 20 cm in total in square it is route 20 which is 10x2 so if its 1.50 dhs per sqm it will be 3 x 10 which is 30 dhs
So 30 x 4 is 120 please mark as brainliest
