I think the problem had a typo.
x^2 - 4x = 4x + 20
it gives imaginary, perhaps the owner of the question meant the right hand side to be 5/(x - 4)
Also, please consider looking for this problem rather than making a new webcalc question because perhaps this question has been posted already and answered already.
These are a system of equations. Add the first two equations, and you get 2a^3 (the 3ab^2 cancel each other out) = 1352. a^3 = 676.
a = cube root of 676.
Plugging back in, we have 676 + 3cbrt(676)b^2 = 679
cbrt(676)b^2 = 1
1/cbrt(676)
b = sqrt(cbrt(676^2)/676) = cbrt(676)/26
a - b = cbrt(676) - cbrt(676) = 25cbrt(676)/26
When looking at the lengths of the sides of the triangle, you realize you can use a certain formula and ignore all the random lines inside the triangle. Heron's formula tells you the area of a triangle given its 3 side lengths.
By heron's formula (which I will not prove and you can look it up on your own), the area of the triangle is \(\sqrt{42(42-36)(42-22)(42-26)} = \sqrt{42*8*20*16} = 32\sqrt{105}\)
wbout the big point theorem :( make the points really big, and they are on the same line. :) especially when you draw a diagram
:) nice job catching my mistake lol.
x^3-4x^2-11x+30
The product is 0 by zero product property when x - 2 = 0, x - 5= 0, x + 3 = 0, so when x = 2 and 5 and -3
A six sided dice has cases: 1, 2, 3, 4, 5, 6.
3 is prime, so you have to get a 1 and a 3 (2 cases)
12 has factors: (1, 12), (2, 6), (3, 4) (6 cases)
Total of 8 working cases. Total number of rolls is 6x6 = 36.
Probability is 8/36 = 2/9
The answer is 134.
Why?
First of all, the question never specified that ACD was not 90 degrees, and without it there would still be sufficient info, but this just makes it easier to visualize things (You can prove that CP = PA without this assumption). Angle CAP is 67 degrees, and let angle PBA be x degrees. We can also construct it so that CD is parallel to AB, and CA makes a 90 degrees angle to both lines. Thus, angle PAB is 90 - 67 = 23 degrees. Since E is collinear with CD, then CE is parallel to AB, and DE is parallel to AB, so shape ABED makes a parallelogram, thus angle ADE is 180 - 23 = 157 degrees. Angle ABE is also 157 degrees, so angle CBE is 157 - angle PBA = 157 - x degrees. Angle CDP is 180 - 157 = 23 degrees, and angle PCD is 90 - 67 = 23 degrees, thus CP = PD, and angle CPD is 180 - 23 - 23 = 134 degrees. By corresponding angles, angle CBE = angle CPD = 157 - x = 134 degrees, and because we are looking for angle CBE, angle CBE = 134 degrees (you dont even really need the x, but its more visuals).
Oops when I wrote CO and OM, I meant CN and NM where N is the intersection of CM and AB, I wonder why I assumed CM intersected O, I think I tricked myself with the diagram, my bad guys. (the answer is still the same lol)
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