The answer is negative 5. Hoped this helps. Please mark me as brainliest if I get it right.
<span>r²sin²θ = 16rcosθ </span>
<span>rsin²θ = 16cosθ </span>
<span>r = 16cosθ / sin²θ </span>
<span>r = 16cotθcscθ</span>
Answer:
y=1/3x_1/5=1/3×15=5_1/5×15=3. 5_3=2
Answer:
x = 3 + √6 ; x = 3 - √6 ;
; 
Step-by-step explanation:
Relation given in the question:
(x² − 6x +3)(2x² − 4x − 7) = 0
Now,
for the above relation to be true the following condition must be followed:
Either (x² − 6x +3) = 0 ............(1)
or
(2x² − 4x − 7) = 0 ..........(2)
now considering the equation (1)
(x² − 6x +3) = 0
the roots can be found out as:

for the equation ax² + bx + c = 0
thus,
the roots are

or

or
and, x = 
or
and, x = 
or
x = 3 + √6 and x = 3 - √6
similarly for (2x² − 4x − 7) = 0.
we have
the roots are

or

or
and, x = 
or
and, x = 
or
and, x = 
or
and, 
Hence, the possible roots are
x = 3 + √6 ; x = 3 - √6 ;
; 