Yea something is not right.if y=6 then you cant solve the system because you end up with x^2-4x-1
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>
If <em>x</em> = -1, you have
2(-1) + 3 cos(-1) + <em>e</em> ⁻¹ ≈ -0.0112136 < 0
and if <em>x</em> = 0, you have
2(0) + 3 cos(0) + <em>e</em> ⁰ = 4 > 0
The function <em>f(x)</em> = 2<em>x</em> + 3 cos(<em>x</em>) + <em>eˣ</em> is continuous over the real numbers, so the intermediate value theorem applies, and it says that there is some -1 < <em>c</em> < 0 such that <em>f(c)</em> = 0.
Answer:
the answer is 396 beacsue i followed yiur instucton on the note answer.
Step-by-step explanation:
