algebra 2 help !!!!!!!!!!!!?

1.

+5 is the y intercept, and -2/3 is the slope, and as it's negative it will be "down"

=> Its y-intercept is at (0, 5) and its slope is down 2 and to the right 3.

2.

-4x + 5y = 7

=> 5y = 7 + 4x

=> y = (1/5)(4x + 7)

=> y = (4/5)x + (7/5)

3.

2x^2 - 3x + 5 = 0

The formula states that:

x = [ -b +/- √(b^2 - 4ac) ] / 2a

In your case, a = 2, b = -3 and c = 5

=> x = [ -(-3) +/- √((-3)^2 - 4(2)(5)) ] / 2(2)

=> x = [ 3 +/- √(9 - 40)] / 4

=> x = [ 3 +/- √-31 ] / 4

=> x = [ 3 + i√31 ] / 4 or x = [ 3 - i√31 ] / 4

4.

gradient = (y1 - y2) / (x1 - x2) = (5 - 2) / (2 - 6) = 3 / -4 = -3/4

thus y = (-3/4)x + c

Where x = 2, y = 5, so 5 = (-3/4)(2) + c

=> 5 = -3/2 + c

=> c = 5 + 3/2 = 13/2

So the equation of this line is y = (-3/4)x + (13/2)

5.

Perpendicular lines have reciprocal gradients

2x - 3y = 6

=> -3y = 6 - 2x

=> y = (1/-3)(6 - 2x)

=> y = -2 + (2/3)x

So the gradient of the perpendicular will be 1/(2/3) = 3/2

y = (3/2)x + c where x=1 and y=2

=> 2 = (3/2)(1) + c

=> c = 2 - (3/2) = (1/2)

So the equation of the perpendicular is y = (3/2)x + (1/2)