Can anyone me with this geometry problem?

  H.  13

Step-by-step explanation:

Make use of the Pythagorean theorem twice. The first time, use it to find TS. The second time, use it to find QR.

  TS² + RT² = RS²

  TS² = RS² -RT² = 64 -48 = 16 . . . . subtract RT², fill in numbers

  TS = 4 . . . . . take the square root

Now, QT = QS -TS = 15 -4 = 11, so ...

  QR² = QT² +RT²

  QR² = 11² +(4√3)² = 121 +48 = 169

  QR = √169 = 13

The length of QR is 13.

step-by-step explanation:

wrgasadvauoyfaiua-00uhf

f(x) = x^3 -2x^2 -3x

x( x^2 -2x-3)

x( x^2 +x -3x -3)

x( x(x+1) -3(x+1))

x( x-3)( x+1)

so zeroes are 0, 3,-1

its (2) option