1) solve 6x^2-13x+7=0
A) 1, 7/6 B) -1,-7/6 C) -6/7,1 D) 1/2, 7/3
2) a rock is thrown off a bridge into a river. It's height, h meters, t seconds after release is given by h= -4.9t^2+6t+13. How long does it take to hit the water.
A) 0.90s B) 1.12s C) 1.22s D) 2.35s
3) solve (3x+6)(2x-4)=(5x-11)(x+2)
A) -2,-1 B) 1,2 C) -2,1 D) -1
A) 1, 7/6 B) -1,-7/6 C) -6/7,1 D) 1/2, 7/3
2) a rock is thrown off a bridge into a river. It's height, h meters, t seconds after release is given by h= -4.9t^2+6t+13. How long does it take to hit the water.
A) 0.90s B) 1.12s C) 1.22s D) 2.35s
3) solve (3x+6)(2x-4)=(5x-11)(x+2)
A) -2,-1 B) 1,2 C) -2,1 D) -1
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1) 6x^2-13x+7=0
(6x-7)(x-1) = 0
x = 7/6 and 1
Answer A)
2) h= -4.9t^2+6t+13 = 0 when it hits the water
Quadratic formula
t = [-6±sqrt(36+254.8)]/-9.8
t = 2.35 seconds
Answer D)
3) (3x+6)(2x-4)=(5x-11)(x+2)
6x^2-24 = 5x^2-x-22
x^2+x-2 = 0
(x+2)(x-1) = 0
x = 1 and -2
Answer C)
(6x-7)(x-1) = 0
x = 7/6 and 1
Answer A)
2) h= -4.9t^2+6t+13 = 0 when it hits the water
Quadratic formula
t = [-6±sqrt(36+254.8)]/-9.8
t = 2.35 seconds
Answer D)
3) (3x+6)(2x-4)=(5x-11)(x+2)
6x^2-24 = 5x^2-x-22
x^2+x-2 = 0
(x+2)(x-1) = 0
x = 1 and -2
Answer C)
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wow, umm, for the first one, just factor it and go from there, the second one, theres no h to substitute in for, and the last one, foil, im in algebra ii and thats hard
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1) You can use the quadratic equation to solve for x. Look it up if you need to because it would seem confusing if I tried to write it on here. What I did is factored it out. I saw 6, 13 and 7. I knew 6+7=13 and so that's why i decided to factor. (6x__)(x__) since I know -6+-7=-13 i put that with 6. (6x-7)(x__), we need positive 7 so the last spot goes to -1. (6x-7)(x-1) From here you can set 6x-7=0 and x-1=0 and solve for x. (figure that out yourself, I am not going to give the answer for you, just show you =] )
2) How long does it take to hit the water? We can assume with that that the height will be 0. so 0=-4.9t^2+6t+13. Use the quadratic formula (look it up) and your a=-4.9, b=6, c=13. You should have two answers. Think reasonably, can time be negative (in this case).
3)Factor these out. (3x*2x)+(3x*-4)+(6*2x)+(6*-4) this is for the left side, do this to the right side. Then bring everything to one side so that you have it equal to zero and you can factor it out. This is similar to number one except you have to factor out to find the equation set to 0 then solve for x.
Good luck I hope this helps you to learn HOW to do it, and not just get the answer.
2) How long does it take to hit the water? We can assume with that that the height will be 0. so 0=-4.9t^2+6t+13. Use the quadratic formula (look it up) and your a=-4.9, b=6, c=13. You should have two answers. Think reasonably, can time be negative (in this case).
3)Factor these out. (3x*2x)+(3x*-4)+(6*2x)+(6*-4) this is for the left side, do this to the right side. Then bring everything to one side so that you have it equal to zero and you can factor it out. This is similar to number one except you have to factor out to find the equation set to 0 then solve for x.
Good luck I hope this helps you to learn HOW to do it, and not just get the answer.