Find an equation of the straight line that is perpendicular to the straight line x+2y=5 and that passes through the point (3,7)

y = 2x + 1

Step-by-step explanation:  

Rewrite the equation in standard form:

x+2y=5

2y = -x + 5

y = -(1/2)x + (5/2)

A line perpendicular to this would have a slope that is the negative inverse of the reference line slope, -(1/2), here.  The new slope would be 2.  The y-intercept, (5/2), will change, so we'll just use b for the new y-intercept, and calculate it later:

y = 2x +b

We know that point (3,7) lies on the new line (it is a solution to the equation), so we can use this point to find the new y-intercept, b.

7 = 2*3 + b

b = 1

The perpendicular line is y = 2x + 1, and goes through point (3,7)

answer: -8.2562234e+13

hope it

step-by-step explanation:

x^(1/6) · x^(1/6) · x^(1/6)   ·x^(1/6) = x^(4/6)   since we add the exponents

4/6 can be simplified to 2/3

the simplified form is x^(2/3)