Posted: 9/30/2013 8:41:08 PM EDT
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Can you solve these?
1) Let f(x)=-x-1, h(x)=-x-5 find (f*h)(-7) 2) Let f(x)=-x-1, h(x)=-x+1 find (h*f)(4) 3) Let f(x)=2x-1, h(x)=x-1 find (f*f)(0) I am helping a friend with math, a fellow gun nerd, but he is struggling with math. I bet him at this time of night, arfcom would figure it out before he did (I am getting him through functions right now) The glory of shame :) |
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f(x)*h(x)=(x-1)(x-5)=x^2-6x+5
49+42+5 (x-1)(x+1)=x^2-1 16-1 (2x-1)(2x-1)=4x^2-4x+1 0+0+1 ETA: you can do the algebra then substitute the values in or substitute the values in and just do the arithmetic. The second is kind of lazy and might not get the lesson across. For example, the last problem could be solved as (2(0)-1)(2(0)-1)=(-1)(-1)=1 ETA2: Your notation is a bit sloppy. I think I understood what you meant but the next post down interprets it differently because it's ambiguous. Math should never be ambiguous 
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Quoted:
Can you solve these? 1) Let f(x)=-x-1, h(x)=-x-5 find (f*h)(-7) 2) Let f(x)=-x-1, h(x)=-x+1 find (h*f)(4) 3) Let f(x)=2x-1, h(x)=x-1 find (f*f)(0) I am helping a friend with math, a fellow gun nerd, but he is struggling with math. I bet him at this time of night, arfcom would figure it out before he did (I am getting him through functions right now) The glory of shame :) 1) Let f(x)=-x-1, h(x)=-x-5 find (f*h)(-7) (-x-1)(-x-5)(-7) = (x^2 - 6x +5)(-7) = -7x^2 + 42x - 30 2) Let f(x)=-x-1, h(x)=-x+1 find (h*f)(4) (-x-1)(-x+1)(4) = (x^2-1)(4) = 4x^2-4 3) Let f(x)=2x-1, h(x)=x-1 find (f*f)(0) Trick question. F*F(0) is (2x-1)(2x-1)(0) = 0 The great thing about ARFCOM is if I fucked them up someone will point it out within 5 minutes. Just post math problems, your work, and your answer and it's graded for you. 
Edit: Shit, are those functions times functions or functions of functions? Further Edit: Functions o' functions work: 1) Let f(x)=-x-1, h(x)=-x-5 find (f*h)(-7) (-(-x-5)-1)(-7) = (x +5)(-7) = -7x - 30 2) Let f(x)=-x-1, h(x)=-x+1 find (h*f)(4) (-(-x-1)+1)(4) = (x+1+1)(4) = 4x+8 3) Let f(x)=2x-1, h(x)=x-1 find (f*f)(0) (2(2x-1)-1)(0) = 0 Worked out for grins: (4x-2-1)(0) = (4x-3)(0) = 0 |
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Quoted:
f(x)*h(x)=(x-1)(x-5)=x^2-6x+5 49+42+5 (x-1)(x+1)=x^2-1 16-1 (2x-1)(2x-1)=4x^2-4x+1 0+0+1 ETA: you can do the algebra then substitute the values in or substitute the values in and just do the arithmetic. The second is kind of lazy and might not get the lesson across. For example, the last problem could be solved as (2(0)-1)(2(0)-1)=(-1)(-1)=1 ETA2: Your notation is a bit sloppy. I think I understood what you meant but the next post down interprets it differently because it's ambiguous. Math should never be ambiguous Ha ha, I bet OP's buddy has them figured out by now. |
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Quoted:
f(x)*h(x)=(x-1)(x-5)=x^2-6x+5 49+42+5 (x-1)(x+1)=x^2-1 16-1 (2x-1)(2x-1)=4x^2-4x+1 0+0+1 ETA: you can do the algebra then substitute the values in or substitute the values in and just do the arithmetic. The second is kind of lazy and might not get the lesson across. For example, the last problem could be solved as (2(0)-1)(2(0)-1)=(-1)(-1)=1 ETA2: Your notation is a bit sloppy. I think I understood what you meant but the next post down interprets it differently because it's ambiguous. Math should never be ambiguous I think you missed a negative or two |
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Quoted:
I think you missed a negative or two Quoted:
Quoted:
f(x)*h(x)=(x-1)(x-5)=x^2-6x+5 49+42+5 (x-1)(x+1)=x^2-1 16-1 (2x-1)(2x-1)=4x^2-4x+1 0+0+1 ETA: you can do the algebra then substitute the values in or substitute the values in and just do the arithmetic. The second is kind of lazy and might not get the lesson across. For example, the last problem could be solved as (2(0)-1)(2(0)-1)=(-1)(-1)=1 ETA2: Your notation is a bit sloppy. I think I understood what you meant but the next post down interprets it differently because it's ambiguous. Math should never be ambiguous I think you missed a negative or two I can't find where I did. Where did I mess up? |
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Quoted:
I can't find where I did. Where did I mess up? Quoted:
Quoted:
Quoted:
f(x)*h(x)=(x-1)(x-5)=x^2-6x+5 49+42+5 (x-1)(x+1)=x^2-1 16-1 (2x-1)(2x-1)=4x^2-4x+1 0+0+1 ETA: you can do the algebra then substitute the values in or substitute the values in and just do the arithmetic. The second is kind of lazy and might not get the lesson across. For example, the last problem could be solved as (2(0)-1)(2(0)-1)=(-1)(-1)=1 ETA2: Your notation is a bit sloppy. I think I understood what you meant but the next post down interprets it differently because it's ambiguous. Math should never be ambiguous I think you missed a negative or two I can't find where I did. Where did I mess up? In the original: Quoted:
Can you solve these? 1) Let f(x)=-x-1, h(x)=-x-5 find (f*h)(-7) 2) Let f(x)=-x-1, h(x)=-x+1 find (h*f)(4) 3) Let f(x)=2x-1, h(x)=x-1 find (f*f)(0) I am helping a friend with math, a fellow gun nerd, but he is struggling with math. I bet him at this time of night, arfcom would figure it out before he did (I am getting him through functions right now) The glory of shame :) |
