=−640x10+1280x9+19904x8−40728x7−144488x6+323904x5−162304x4+1024x3+2048x2
step by step
(2x+8x2+3x−4x(x−4)(x−1)(20)x+4)(2)(x+4)(x−4)(2)x(x−1)(2)x(x+4)
=((2x+8x2+3x−4x(x−4)(x−1)(20)x+4)(2)(x+4)(x−4)(2)x(x−1)(2)x)(x+4)
=((2x+8x2+3x−4x(x−4)(x−1)(20)x+4)(2)(x+4)(x−4)(2)x(x−1)(2)x)(x)+((2x+8x2+3x−4x(x−4)(x−1)(20)x+4)(2)(x+4)(x−4)(2)x(x−1)(2)x)(4)
=−640x10+3840x9+4544x8−58904x7+91128x6−40608x5+128x4+512x3−2560x9+15360x8+18176x7−235616x6+364512x5−162432x4+512x3+2048x2
=−640x10+1280x9+19904x8−40728x7−144488x6+323904x5−162304x4+1024x3+2048x2
Answer:
2006
Step-by-step explanation:
1 + 2(1027) + 6 + 7 - 8 • 9 + 10
1 + 2054 + 6 + 7 - 72 + 10
2006
Answer:
50 degrees and 40 degrees
Step-by-step explanation:
50 degrees and 40 degrees
complimentary = add up to 90 degrees
5:4:9
50:40:90
Given
s=16t^2
where
s=distance in feet travelled (downwards) since airborne with zero vertical velocity and zero air-resistance
t=time in seconds after release
Here we're given
s=144 feet
=>
s=144=16t^2
=>
t^2=144/16=9
so
t=3
Ans. after 3 seconds, the object hits the ground 144 ft. below.
