Answer:

Step-by-step explanation:

I got this:
Tennis court is a rectangle
(Rectangle - 2 sides the same length and the other 2 sides the same. So 2 pairs of equal length sides)
If one is 27 the other has got to be 27 as well - 27+27= 54
210 - 54 = 156
156 / 2 = the other "pair"
Hope that helps and correct me if I'm wrong!
Answer:
1) (x + 3)(3x + 2)
2) x= +/-root6 - 1 by 5
Step-by-step explanation:
3x^2 + 11x + 6 = 0 (mid-term break)
using mid-term break
3x^2 + 9x + 2x + 6 = 0
factor out 3x from first pair and +2 from the second pair
3x(x + 3) + 2(x + 3)
factor out x+3
(x + 3)(3x + 2)
5x^2 + 2x = 1 (completing squares)
rearrange the equation
5x^2 + 2x - 1 = 0
divide both sides by 5 to cancel out the 5 of first term
5x^2/5 + 2x/5 - 1/5 = 0/5
x^2 + 2x/5 - 1/5 = 0
rearranging the equation to gain a+b=c form
x^2 + 2x/5 = 1/5
adding (1/5)^2 on both sides
x^2 + 2x/5 + (1/5)^2 = 1/5 + (1/5)^2
(x + 1/5)^2 = 1/5 + 1/25
(x + 1/5)^2 = 5 + 1 by 25
(x + 1/5)^2 = 6/25
taking square root on both sides
root(x + 1/5)^2 = +/- root(6/25)
x + 1/5 = +/- root6 /5
shifting 1/5 on the other side
x = +/- root6 /5 - 1/5
x = +/- root6 - 1 by 5
x = + root6 - 1 by 5 or x= - root6 - 1 by 5
Equation 1: y = -2x + 1
Equation 2: y = 2x - 3
Since both equations already have y isolated, we are able to simply set the right side of both equations equal to each other. Since we know that the value of y must be the same, we can do this.
-2x + 1 = 2x - 3
1 = 4x - 3
4 = 4x
x = 1
Then, we need to plug our value of x back into either of the original two equations and solve for y. I will be plugging x back into equation 2 above.
y = 2x - 3
y = 2(1) - 3
y = 2 - 3
y = -1
Hope this helps!! :)
Only D is correct, it intersects at 90 degrees so it cant be obtuse or acute and never intersecting would be parallel lines <span />