What is the y-intercept of the equation of the line that is perpendicular to the line y=3/5x+10 and passes through the point (15,-5)? A. y=3/5x-20 B. y=-3/5x+20 C. y=5/3x-20 D. y=-5/3x+20

10 is the y intercept

Step-by-step explanation:

follow y=mx+b b is equal to the y intercept

D. y=-5/3x+20

Step-by-step explanation:

We have the information y=3/5x+10, the lines are perpendicular, and the new line passes through (15,-5). The slops of perpendicular lines are negative reciprocals so you need to find the negative reciprocal of 3/5, so flip it to 5/3 and multiply by -1, we get the slope of the new line as -5/3. So far we have the equation y=-5/3x+b. We are given a point on the line, (15,-5), we can plug these into the equation as x and y to solve for the y-intercept, b. You set it up as -5=-5/3(15)+b. First you multiply to get -5=-25+b, then you add 25 to both sides to isolate the variable and you get b=20. Then you can use b to complete your equation with y=-5/3x+20, or letter D.

it would be 6xy and -16xy.

answer: i think it is (1,-2)