Let
x----> the length side of the hypotenuse
y----> the length side of the <span>unknown leg

we know that
x=2*y----> equation 1
applying the Pythagoras Theorem
x</span>²=y²+3²-----> equation 2

substitute equation 1 in equation 2
[2*y]²=y²+3²----> 4*y²-y²=9-----> 3*y²=9----> y²=3
y=√3 ft
x=2*y----> x=2*√3 ft

the answer is
<span>the lengths of the three sides are
</span>hypotenuse=2√3 ft
one leg=√3 ft
other leg=3 ft

7 0

3 years ago

Decide, without calculation, if each of the integrals below are positive, negative, or zero. Let D be the region inside the unit

Schach [20]

The integrals over B and T will be positive. Keeping $y$ fixed, $xe^x$ is strictly increasing over D as $x$ increases, so the integrals over $x$ (i.e. the bottom/top left quadrants of D) is negative but the integrals over $x>0$ are *more* positive.

The integrals over R and L are zero. If we take $f(x,y)=xe^x$ , then $f(x,-y)=f(x,y)$ , which is to say $f$ is symmetric across the $x$ -axis. For the same reason, the integral over all of D is also zero.

4 0

3 years ago