For line B to AC: y - 6 = (1/3)(x - 4); y - 6 = (x/3) - (4/3); 3y - 18 = x - 4, so 3y - x = 14
For line A to BC: y - 6 = (-1)(x - 0); y - 6 = -x, so y + x = 6
Since these lines intersect at one point (the orthocenter), we can use simultaneous equations to solve for x and/or y:
(3y - x = 14) + (y + x = 6) => 4y = 20, y = +5; Substitute this into y + x = 6: 5 + x = 6, x = +1
<span>So the orthocenter is at coordinates (1,5), and the slopes of all three orthocenter lines are above.</span>
d=28g I might be wrong, but we can find this by dividing 140 by five, which equals 28.
Answer:
Cuando rechazamos la hipótesis nula, esperaríamos que la razón de varianza, a largo plazo, sea: <u>mayor al valor crítico.</u>
Step-by-step explanation:
That's OK, but you have not said which variable you want to solve it for.
<u>To solve for 'x':</u>
<span>c + ax = dx
Subtract c from each side: ax = dx - c
Subtract dx from each side: ax - dx = -c
Factor the left side: x (a - d) = -c
Divide each side by (a - d) : x = -c / (a - d) or <u>x = c / (d - a)</u> .
</span><span><u>To solve for 'c': </u>
</span><span> c + ax = dx
Subtract ax from each side and factor: <u>c = x (d - a) </u>
</span><u>To solve for 'd': </u>
<span>c + ax = dx
Divide each side by 'x': d = c/x + a .
<u>To solve for 'a':</u>
</span><span><span> c + ax = dx</span>
Subtract 'c' from each side: ax = dx - c
Divide each side by 'x': <u>a = d - c/x </u>.
.</span>
Answer:
addition
Step-by-step explanation:
if u add 5 to both sides, the -5 and the +5 would cancel out, leaving you with x=17
