Where are the statements please?
Sorry this is really sloppy but hope this helps!
Answer:
x+3y-6=0
Step-by-step explanation:
given eqn is y=3x-2 which is 3x-y-2=0
the eqn of line perpendicular to given eqn is -x+3y+k=0
it passes through (6,4)
-6+3*4+k=0
or,. -6+12+k=0
or, k= -6
therefore, the eqn of line perpendicular to given eqn is x+3y-6=0
x = 2(180 - x) - 12
3x = 348
x = 116
The measure of the angle is 116 degrees.
Answer:heyyy there...the answer is b that is -12x^3+12x^2-20
Step-by-step explanation:
<h3>f(x)=
-12x^3+19x^2-5</h3><h3>g(x)=
7x^2+15</h3><h3>f(x)-g(x)=(
-12x^3+19x^2-5)-( 7x^2+15)...{while opening the bracket the sign of the second polynomial changes accordingly}</h3><h3>
it becomes -12x^3+19x^2-5-7x^2-15</h3><h3>
=-12x^3+12x^2-20</h3><h3>
HOPE IT HELPED UUUU</h3>
