Hi
k (a+b) = ka+kb
so : 8 (3x+4) = 8*3x +8*4 = 24x +32
-4 is the x-intercept for your equation
The <em>xy</em>-plane has a normal vector of 〈0, 0, 1〉, and any plane parallel to it will have the same normal vector.
Then the equation of the plane through (6, 3, 2) that is parallel to the <em>xy</em>-plane has equation
〈<em>x</em> - 6, <em>y</em> - 3, <em>z</em> - 2〉 • 〈0, 0, 1〉 = 0
==> <em>z</em> - 2 = 0
==> <em>z</em> = 2
Use: (a + b)(c + d) = ac + ad + bc + bd
Answer:
d. x² + 6x +8
Okay, we start with 42x^2 + 32 - 18 = 6062. To solve single-variable equations like this one, we want to isolate the x by moving all of the other numbers to the other side of the equation. We need to first subtract 32 and 18, and that equals 14. Now our problem looks like this, 42x^2 + 14 = 6062. Since 14 is the only other number on the left side, we subtract four from both sides! This gives us 42x^2 + 14 - 14 = 6062 - 14, so 42x^2 = 6062. Now to find x, we want to just have one x on the left side instead of 42, so we divide the equation by 42 to find that 42x^2/42 = 6048/42, so x = 12
