Question 757719
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Plot your problem on a set of coordinate axes, assuming without loss of generality, that the vertical wall is co-incident with the positive *[tex \LARGE y]-axis, the ground is co-incident with the *[tex \LARGE x]-axis, and the origin, *[tex \LARGE (0,0)] is the point of intersection with the wall and the ground.


Given that the positive direction in the *[tex \LARGE x]-axis is from the base of the ladder toward the wall, the base of the ladder must be at the point *[tex \LARGE (-4,0)].  You are given that the slope of the ladder is 3.


Use the point-slope form of an equation of a line to derive an equation of the line that contains the line segment representing the ladder.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ y_1\ =\ m(x\ -\ x_1) ]


where *[tex \Large \left(x_1,y_1\right)] are the coordinates of the given point and *[tex \Large m] is the given slope.


Since the ladder touches the wall at some point *[tex \LARGE (0,b)], substitute zero for *[tex \LARGE x] in your equation and solve for the value of *[tex \LARGE b] which will be the vertical height of the top of the ladder.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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