No. It would be negative 5.
Answer:
x=5
:)
Step-by-step explanation:
Answer:
one third :) ;) hope it helps
<span>Use a straightedge to join points W and P and then points P and X. â–łWPX is equilateral.
Let's see now, Delmar has a line segment WX and has drawn 2 circles whose radius is the length of WX, centered upon W and centered upon X. Sounds to me that all he needs to do is select one of the intersections of those 2 circles and use that at the 3rd point of the equilateral triangle and draw a line from that point to W and another line from that point to X. Doesn't matter which of the two intersections he chooses, just needs to pick one. Looking at the available options, only the 1st one which is "Use a straightedge to join points W and P and then points P and X. â–łWPX is equilateral." matches my description, so that is the correct choice. The other choices tend to do rather bizarre things like create a perpendicular bisector of WX and for some unknown reason, claim that bisector is somehow a side of a desired equilateral triangle.</span>
Step-by-step explanation:
- Assume non-hypotenuse sides are 4 and 5, so tan =4/5=0.8 . Then hypotenuse is sqrt(16+25) = sqrt(41) So sides are 4,5,sqrt (41).
