Rewrite <span>2cos x + 1 = 0 as:
2 cos x = -1, and then cos x = -1/2
x must be in Quadrant II or Quadrant III, since the adj. side is negative.
Note that the angle 120 has adj. side -1 and hyp 2. So 120 degrees is one solution.
Now what about a possible 2nd solution, to be found in Quadrant III? That would be -120 degrees, which has the same terminal line as does 240 degrees.
No soap.
So, the solution is 120 degrees.</span>
Answer:
x=6
Step-by-step explanation:
81^x = 27 ^(x+2)
81 = 3^4 and 27 =3^3 so replace 81 and 27 in the equation
3^4^x = 3^3^(x+2)
When we have a power to a power we can multiply the exponents
a ^b^c = a^(b*c)
3^(4x) = 3^(3*(x+2))
Since the bases are the same, the exponents have to be the same
a^b = a^c means b=c
4x = 3(x+2)
Now we can solve for x
Distribute
4x = 3x+6
Subtract 3x from each side
4x-3x = 3x-3x+6
x = 6
Answer:
1,215 i think please be right
vertex = (3,- 5 )
given a quadratic in standard form : y = ax² + bx + c ( a ≠ 0 ), then
the x-coordinate of the vertex is
= - 
y = x² - 6x + 4 is in standard form
with a = 1, b = - 6 and c = 4, hence
= - 
= 3
substitute this value into the equation for y- coordinate
y = 3² - 6(3) + 4 = 9 - 18 + 4 = - 5
vertex = (3, - 5 ) → second table
