Can someone me ?
joshua has a ladder that is 11 ft. long. he will lean the ladder against a vertical wall. for safety reasons, he wants the angle the ladder makes with the ground to be no greater than 65°. is it possible for joshua to lean the ladder against the wall so that the top of the ladder is at least 10 ft. above the ground?

Step-by-step explanation:

It's close, but not close enough to be at least 10 feet above the ground.  This is a classic right triangle trig problem.  We have the length of the ladder, which is the hypotenuse of the right triangle, to be 11 ft.  The base angle, or the angle of inclination, is 65.  We are looking for the height up the wall that the ladder will go at this angle.  The height of the wall is the side opposite the base angle, so the sin ratio applies here.

and

11 sin(65)= y so

y = 9.96938 ft

Again, close but not quite there.

answer rather than having to add and subtract the measurements of the semi-circle, notice that the semicircle to the left of ad is actually identical in area to the blank semicircle-shaped space to the left of bc. so if we transferred the red semicircle into the blank space, we would end up with a square of side length 24 cm. the area of a square is s^2 = 24^2 = 676 cm^2.

u also answered this so i think u would know but i am here to

put in or use mine

because linear equations are a lot simpler than quadratics equations, and linear equations formula is a lot easier to memorize. that's why i think they're harder.