the net is pretty much the net of a long box, kinda like the one in the example in the picture below. Due to that, we can pretty much assume the two sides sticking up and down, are just two small 6x3 rectangles, namely, they have a height of 3, reason why we assume that, if that if we fold the other sides to make out the box, those two sides sticking out, must be 3m to neatly snugfit.
now, if we close the box as it stands, the sides(laterals) will be on the left-right sides two 3x15 rectangles, and on the front-back sides, two 6x15 rectangles.
we're excluding the top and bottom sides, because those are not "laterals", or sides of the box.
Error: They subtracted 9x from both sides when they should have subtracted 12. You can’t subtract unlike terms
Correct: x=-5/16
Subtract 12 from both sides to get -5 on the left and add 7x and 9x to get 16x. Divide both sides by 16 to isolate x.
Given:
Arc(AB) = 78 degrees
Measure of angle CMD = 106 degrees
To find:
The measure of arc CD.
Solution:
Secant intersection theorem: If two secant of a circle intersect each other inside the circle, then the intersection angle is the average of intercepted arcs.
Using secant intersection theorem, we get
Multiply both sides by 2.
Therefore, the measure of arc CD is 134 degrees and the correct option is C.
Hello from MrBillDoesMath!
Answer:
Case 1: x is a real number. In this case f(x) has no real solution.
Case 2: x is a complex number. In this case the roots of f(x) are
x = 3 +\- 2 sqrt(3) i
(sqrt = square root. "i" = sqrt(-1))
Discussion:
Add 1 to both sides of the given equation:
2(x-3)^2 -1 + 1 = -25 + 1 or
2(x-3)^2 + 0 = -24 or
2(x-3)^2 = -24 (*)
Case 1: x is a real number. In that case as the square (x-3)^2 is always >0, so (*) has no real solution.
Case 2. x is a complex number ( a + bi, where a <> 0). In this case divide both sides of (*) by 2,
(x-3)^2 = -24/2 = -12 =>
x-3 = +\- sqrt(-12) = +\- sqrt (12 * -1) = +\- sqrt (4 * 3 * -1) = +\- 2 sqrt(3) * i
where i = sqrt (-1).
Add 3 to both sides gives
x = 3 +\- 2 sqrt(3) i
Thank you,
MrB
