Algebra 2 Homework . . . Desperate!?
I feel really selfish asking for this :-/
I'm at a dress rehearsal for my school musical right now and we'll be here until at least eleven at night. It's the last one before the show Friday and I have math homework that I have no time to do!
I was hoping somebody on Yahoo! could help me. I pretty much know the material . . . but like I said, being in musical and probably not getting home until late at night, I won't be able to complete it. It's only 16 problems, and they're not too hard.
I'm asking you guys if you could help me out a whole bunch and do as much as you can, showing your work and answers!
Again, I feel selfish asking, but I am desperate!
Here's the problems:
1. Given the points P(3, -5) and Q(6, 1) find (a) the length of PQ, (b) the midpoint of the segment PQ, and (c) the point X such that P is the midpoint of QX.
2. Find an equation of the circle having center (3, 3) and containing the origin.
3. Find the center and radius of the circle x^2 + y^2 - 10x = 0
4. Find an equation for the set of all points in the plane equidistant from the point (2, 3) and the line x = 4.
5. Find the vertex, focus, directrix, and axis of the parabola (y - 3)^2 = 7( x - 5).
6. Find an equation of the ellipse having foci ( negative square root of 7 , 0 ) and ( square root of 7 , 0 ) and sum of focal radii 8.
7. Grapph the ellipse 16 x^2 + 4y^2 = 16.
8. Graph the hyperbola y^2 - x^2 + 9 = 0. Show the asymptotes as dashed lines.
9. Graph the circle. Label the center and the radius.
x^2 + y^2 - 4x + 2y +1 = 0
10. Graph the parabola: x = 1/4 ( y + 1 )^2 + 2.
Find all real number solutions for the following systems:
11. x^2 + y^2 = 5
2x^2 + 3y^2 = 11
12. 4x^2 + y^2 = 25
2x + y = 1
13. Solve the system: x^2 + y^2 = 25
2x + y = 10
14. Find the dimensions of a rectangle having perimeter 28 cm and a diagonal of length 10 cm.
15. Solve the system: 2x - 5y + 4z = 18
x + 3y + 5z = 7
-3x - 2y + 4z = 22
16. x + y - z = 2
3x + 2y - z = 5
5x + 2y + z = 7
Again, thank you so much!
- 7 years agoFavorite Answer
1a. to find the distance between the x & y coordinates, find the absolute value of x1-x2 and absolute value of y1-y2. then use Pythagorean.
So (|3-6|)^2+ (|-5-1|)^2=c^2 --> 3^2+6^2=c^2 --> c=6.708...
1b. to find the midpoint, the x coordinate is (x1-x2)/2 and the y coordinate is (y1+y2)/2.
So the midpoint is ( 3+6/2, -5+1/2) --> (4.5, -2)
2. to find the equation, plug in numbers into the equation (x - h)² + (y - k)² = r². (h,k) is the center coordinates and to find r, you need to plug in the "pass through" coordinates for x & y. (x,y)=origin=(0.0)
So (0 - 3)² + (0 - 3)² = r² -->9+9 = r² --> 18=r² So the equation is (x - 3)² + (y -3)² =18
3. so you need to make the equation look like (x - h)² + (y - k)² = r² and in order to do that you need to complete the square for x and find the values. How to complete the square-http://www.mathsisfun.com/algebra/completing-squar...
so (x-5)^2-25 +y^2=0 --> (x-5)^2+(y-0)^2=25 --> r=radius=5 and center=(5,0)
5. manipulate the equation to (x - h)² = 4p(y - k) but it is difficult/impossible with the equation you provided so i don't know.
6. so translate foci to (-c,0) and (c,0) and sum to 2*a and b^2=a^2-c^2
making c=sqrt 7 and a=8/2=4, solving for b you get 3. plug into the formula (x^2/a^2)+(y^2/b^2)=1
radius of 2, center= (2,-1)
and sorry, I have to stop, my laptop is dying.Source(s): math tutor