course Mth 271 This is the rest of assignment 4 ????? ??????assignment #004{???H??y??????
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14:53:14 #22. What value of b would we use to express various functions in the form y = A b^x? What is b for the function y = 12 ( e^(-.5x) )?
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RESPONSE --> e^-,5= .607 y= 12* .607^x b= .607
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14:53:55 ** 12 e^(-.5 x) = 12 (e^-.5)^x = 12 * .61^x, approx. So this function is of the form y = A b^x for b = .61 approx.. **
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RESPONSE --> I understand this my answer is the same but I rounded a 3 places instead of 2.
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14:54:54 what is b for the function y = .007 ( e^(.71 x) )?
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RESPONSE --> e^.71= 2.034 b= 2.034 y= .007 * 2.034^x
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14:55:39 ** 12 e^(.71 x) = 12 (e^.71)^x = 12 * 2.04^x, approx. So this function is of the form y = A b^x for b = 2.041 approx.. **
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RESPONSE --> I understand this an seem to have gotten the correct answer
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14:56:25 what is b for the function y = -13 ( e^(3.9 x) )?
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RESPONSE --> e^3.9= 49.4 b= 49.4 y= -13(49.4)^x
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14:56:56 ** 12 e^(3.9 x) = 12 (e^3.9)^x = 12 * 49.4^x, approx. So this function is of the form y = A b^x for b = 49.4 approx.. **
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RESPONSE --> I seem to understand this material pretty good. I am getting the correct answers
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14:58:20 List these functions, each in the form y = A b^x.
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RESPONSE --> opps! I guess I should have waited to do that....... y= 12*.607^x y= .007*2.034^x y= -13*49.4^x
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14:58:57 ** The functions are y=12(.6065^x) y=.007(2.03399^x) and y=-13(49.40244^x) **
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RESPONSE --> On these it looks like you rounded off further than me.....but it looks like we both have the same idea
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14:59:52 0.4.44 (was 0.4.40 find all real zeros of x^2+5x+6
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RESPONSE --> it breaks up to (x+3)(x+2) zeros are at x=-3 and x= -2
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15:00:48 ** We can factor this equation to get (x+3)(x+2)=0. (x+3)(x+2) is zero if x+3 = 0 or if x + 2 = 0. We solve these equations to get x = -3 and x = -2, which are our two solutions to the equation. COMMON ERROR AND COMMENT: Solutions are x = 2 and x = 3. INSTRUCTOR COMMENT: This error generally comes after factoring the equation into (x+2)(x+3) = 0, which is satisfied for x = -2 and for x = -3. The error could easily be spotted by substituting x = 2 or x = 3 into this equation; we can see quickly that neither gives us the correct solution. It is very important to get into the habit of checking solutions. **
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RESPONSE --> I should have shown more work I think.......I didnt show that I set them both = to zero and then solved but I do understand how this works
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15:01:45 Explain how these zeros would appear on the graph of this function.
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RESPONSE --> They would be at (-2,0) and at (-3,0) .... when x=-2 or x=-3 then y=0
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15:02:24 ** We've found the x values where y = 0. The graph will therefore go through the x axis at x = -2 and x = -3. **
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RESPONSE --> I actually worded this last one pretty close to yours so I think it is a pretty good answer
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15:03:55 0.4.50 (was 0.4.46 x^4-625=0
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RESPONSE --> Sqrt(625)= 25 (x^2+25) (x^2-25) x= -5 and x= 5 for both
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15:04:46 ** Common solution: x^4 - 625 = 0. Add 625 to both sides: x^4 = 625. Take fourth root of both sides, recalling that the fourth power is `blind' to the sign of the number: x = +-625^(1/4) = +-[ (625)^(1/2) ] ^(1/2) = +- 25^(1/2) = +-5. This is a good and appropriate solution. It's also important to understand how to use factoring, which applies to a broader range of equations, so be sure you understand the following: We factor the equation to get (x^2 + 25) ( x^2 - 25) = 0, then factor the second factor to get (x^2 + 25)(x - 5)(x + 5) = 0. Our solutions are therefore x^2 + 25 = 0, x - 5 = 0 and x + 5 = 0. The first has no solution and the solution to the second two are x = 5 and x = -5. The solutions to the equation are x = 5 and x = -5. **
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RESPONSE --> I did this the second way because I think that way is the easiest......the answers seem to match up .....
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15:06:58 0.4.70 (was 0.4.66 P = -200x^2 + 2000x -3800. Find the x interval for which profit is >1000
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RESPONSE --> 1000= -200x^2 + 2000x- 3800 0= -200x^2+ 2000x - 4800 x^2-10+24 (X-6)(X-4) so x=6 and x= 4
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15:08:18 ** You have to solve the inequality 1000<-200x^2+2000x-3800, which rearranges to 0<-200x^2+2000x-4800. This could be solved using the quadratic formula, but it's easier if we simplify it first, and it turns out that it's pretty easy to factor when we do: -200x^2+2000x-4800 = 0 divided on both sides by -200 gives x^2 - 10 x + 24 = 0, which factors into (x-6)(x-4)=0 and has solutions x=4 and x=6. So -200x^2+2000x-4800 changes signs at x = 4 and x = 6, and only at these values (gotta go thru 0 to change signs). At x = 0 the expression is negative. Therefore from x = -infinity to 4 the expression is negative, from x=4 to x=6 it is positive and from x=6 to infinity it's negative. Thus your answer would be the interval (4,6). COMMON ERROR: x = 4 and x = 6. INSTRUCTOR COMMENT: You need to use these values, which are the zeros of the function, to split the domain of the function into the intervals 4 < x < 6, (-infinity, 4) and (6, infinity). Each interval is tested to see whether the inequality holds over that interval. COMMON ERROR: Solution 4 > x > 6. INSTRUCTOR COMMENT: Look carefully at your inequalities. 4 > x > 6 means that 4 > x AND x > 6. There is no value of x that is both less than 4 and greater than 6. There are of course a lot of x values that are greater than 4 and less than 6. The inequality should have been written 4 < x < 6, which is equivalent to the interval (4, 6). **
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RESPONSE --> I am in the common error catagory.......I just didnt think things though.....I understand that there is a gap between 4 and 6 and that at 4 it goes to - infinity and at 6 it go to + infinity
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