Answer:
Step-by-step explanation:
I don’t know
there will be one solution because the lines intersect at exactly one set of points.
(0,6)(6,3)
slope(m) = (3-6) / (6-0) = -3/6 = -1/2
y = mx + b
6 = -1/2(0) + b
6 = b
y = -1/2x + 6
(0,0)(6,6)
slope(m) = (6-0) / (6-0) = 6/6 = 1
y = mx + b
6 = 1(6) + b
6 = 6 + b
6 - 6 = b
0 = b
y = x + 0
x = -1/2x + 6
1/2x + x = 6
1/2x + 2/2x = 6
3/2x = 6
x = 6/(3/2)
x = 6 * 2/3
x = 12/3 = 4
<span>solution is : (4,4)</span>
Answer:
Step-by-step explanation:
This is an interesting question. I chose to tackle it using the Law of Cosines.
AC² = AB² + BC² - 2·AB·BC·cos(B)
AM² = AB² + MB² - 2·AB·MB·cos(B)
Subtracting twice the second equation from the first, we have
AC² - 2·AM² = -AB² + BC² - 2·MB²
We know that MB = BC/2. When we substitute the given information, we have
8² - 2·3² = -4² + BC² - BC²/2
124 = BC² . . . . . . . . . . . . . . . . . . add 16, multiply by 2
2√31 = BC ≈ 11.1355