  # dcowboys107

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## dcowboys107's Achievements 10

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1. (1-x^2)^(1/2) -------------- (4-x^2)^(1/2) One minus x squared divided by 4 minus x squared all under a square root sign. My teacher says the domain is (-inf,-2) union [-1,1] union (2,inf) Is it necessary to include the [-1,1]? Wouldn't it be obvious that function is valid at that point? I have a tough time deciding when to include those terms How do I know when to include the infinity and such?
2. I don't think we used L'Hopital or something. Is their a different way?
3. I have some limit problems and I'm not sure how to do it. I already have the answers so I'm not asking for answers but the "how to". lim (1/x)-(1/3) ----------- x--->3 (x-3) lim (t+4)^(1/2)-2 t-->0 ----------- t I always end up getting 0/0 on all the fractions. I'm not sure what to do for these. Thanks!
4. I think I got it to work. Does that seem like the right answer though?
5. Write equations for the two straight lines that pass through the poing (2,5) and are tangent to the parabola y=4x-x^2. I drew the figure and the lines. I fought the slope for the tangent lines (2 and -2) and eventually the equations y=-2x+9 and y=2x+1 When I plugged the three equations into my calcuator to see if I was right it seems that I was off on y=-2x+9, that it doesn't reach 5 when x is two but at 4. The second one seems right. Am I right and the calcultar be wrong? Thanks.
6. From a point on level ground, an observer measures the angle of elevation to the top of a hill to be 38 degrees The observer then walks 370 meters directly away from the hill and measures the angle of elevation to the top of the hill to be 25 degrees. Determine the height of the hill to the nearest meter. I drew it out and for variables I used "x" as the distance from the first angle measurement to the base of the hill and 370 from the end of the first measurement to the start of the second measurement. I used "h" to represent height. I got cot 38 degrees=x/h and cot 25=(370+x)/h I solved for x in the first equation then plugged it into the second and got finally h=370/(cot 25 - cot 35) why is this answer wrong? Thanks for the help!
7. Rewrite f(x) = 4csc^2(x) + 9tan^2(x) in terms of cos(x). NOTE: Expressions such as cos^2(x) must be entered as cos(x)^2 or (cos(x))2. I rewrote csc(x)^2 as 1/1-cos(x)^2 and tan^2 as ((1-cos(x)/cos(x))^2 I'm having a hard time simplifying that down to just cos(x).
8. What is the value of x for these equations? a. Find all solutions for this equation: (e^2x)-(6e^x)+32/4=0 b.(e^x)-24e^(-x)=-2 Use common logs to solve for x in terms of y for the following: c.y=10^x+10^(-x) ----------------- 18 I don't know even where to start. . . I'd apprecite all the help possible.
9. I'm still not getting the answer. . . could you show me step by step?
10. a.ln(x)-ln(6-x)=3 b.log base 2 of (x+5)=log base 2 of (x-5)+log base 6 of (36)+6^(log base 6 of 3) a.ln(x) - ln(6-x) = 3 ln(x/(6-x)) = 3 x/(6-x) = e³ I've gotten this far for part a but now I can't solve for x. log[base 2] ((x+5)/(x-5)) = 2 + 3 (x+5)/(x-5) = 2^5 Part b I've gotten this far. It says not to give a decimal approximation for any of the problems rather a fraction or an exact expression. Can someone show me how to solve for x in these two problems? Thanks!
11. I got -3/10 by using -b/2a. I put it into the expression and got said number. How do you find the what the value is then?
12. The graph of a quadratic function f(x) has its vertex at (5,-6) and passes through the point (2,-51). Find a formula for f(x) I got 15(x-5)^2 - 6 Why is this wrong? Find the minimum value of the function (5x-2)(2x+2). I got -49/20 after multiplying out the equation to get 5x^2+3x-2 (simplified). I did -3/10 and put it back into the equation and got it wrong. . . Any ideas?
13. We have to enter our answers online and it says I've gotten the wrong answer. . . How do I enter the answer in? I'm very confused becuase I feel like I've done it right. . .
14. It has been estimated that 1000 curies of a radioactive substance introduced at a point on the surface of the open sea would spread over an area of 20,000 km2 in 20 days. Assuming that the area covered by the radioactive substance is a linear function of time t and is always circular in shape, express the radius r of the contamination as a function of t. Apparently Area = constant * t. As 20,000 km2 = constant * 20 days, the constant equals 1,000 km2/day Therefore Area = 1,000 km^2/day * t On the other hand Area = pi r^2. So pi r^2 = 1,000 km^2/day * t, meaning that radius r = sqrt (1000/pi km^2/day) * t^(1/2) I got this far but I don't think the answer has km as part of the solution. I'm sure that there is a pi involved. Any ideas?
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