<span>Assume that,
x^2+4x-5 = 0 .......(1)
Then,
x^2+4x-5 = 0
x^2+5x-1x-5 =0
x(x+5)-1(x+5) = 0
(x+5) (x-1) = 0
We get x=-5 and x=1
Sub x=-5 in equ (1)
(-5)^2+4(5)-5 = 0
-25+20-5 = 0
-25+25= 0
0 = 0
Sub x=1 in equ (1)
(1)^2+4(1)-5 = 0
1+4-5 = 0
5-5 = 0
0 = 0
Therefore x value is -5 and 1</span>
Answer:
(a) The unit circle is centered at (0,0) with a radius of 1.
(b) The equation of a circle of radius <em>r</em>, with a center located at (0,0):
<em>x</em>²<em>+ y</em>² <em>= r</em>².
(c) (i) P(1,0)
(ii) P(0,1)
(iii) P(-1,0)
(iv) P(0,-1)
Step-by-step explanation:
Answer:
I think the answer is F hope its right fingers crossed!
Step-by-step explanation:
2r + 2r + 5 + r + 4r - 3 = 38
9r + 5 - 3 = 38
9r + 2 = 38
9r = 38 - 2
9r = 36
9r/9 = 36/9
r = 4cm
Done!
Answer:
The mid-point between the endpoints (10,5) and (6,9) is:
Step-by-step explanation:
Let (x, y) be the mid-point
Given the points
Using the formula to find the mid-point between the endpoints (10,5) and (6,9)
Here:
Thus,
Therefore, the mid-point between the endpoints (10,5) and (6,9) is:
