2 ft 7 in
3 ft 11 in
6 ft 9 in
-----------------add
11 ft 27 in........and since 1 ft = 12 inches, then 27 inches = 2 ft 3 inches
so it would become : 13 ft 3 inches <==
Draw points J and K in the coordinate axis, and join them so that the top of the vector is point K.
the magnitude of the vector is just the length of the segment JK, so we can either write it as |JK| or |KJ|, it does not make any change.
Let O be the point in the x axis such that KO is perpendicular to OJ, as shown in the picture.
From the Pythagorean theorem:

units
since |OK|=|OJ|, the right triangle KOJ is isosceles, so the measure of KJO is 45°, which means that the angle of vector JK to the positive x axis is 180°-45°=135°
Answer: magnitude=

units, direction : 135°
Try www.tigeralgebra.com it helps me all the time
20 pounds of lollipops. 20 x .95 = 19, 10 x 1.1 = 11. sorry i'm awful at explaining math.
<span>1. circle
2. parabola
3. ellipse
4. hyperbola
Let's look at each equation and see what they are:
1. x^2 + y^2 - 4x + 6y - 5 = 0
* There's a lot of crud in this equation, but the thing to note is that the x^2 and y^2 terms have the same scaling factor (which is 1). This should scream "circle" to you.
2. x^2 - 6y = 0
* Key thing to note here is that the y term isn't squared, but the x term is squared. This is a key sign that the equation is a parabola.
3. 4x^2 + 9y^2 = 1
* Here we have the sum of an x squared term and a y squared term. That kinda sounds like a circle, but there's those 2 coefficients scaling the results. And they're different. So we're looking for a closed curve that kinda looks like a circle, but it's stretched out a bit. And that's an ellipse.
4. 7x^2 - 9y^2 = 343
* Here we have a couple of squared terms for x and y. But we're not adding them together, we're subtracting. And that indicates a hyperbola.</span>