the parallel line is 2x+5y+15=0.
Step-by-step explanation:
ok I hope it will work
soo,
Solution
given,
given parallel line 2x+5y=15
which goes through the point (-10,1)
now,
let 2x+5y=15 be equation no.1
then the line which is parallel to the equation 1st
2 x+5y+k = 0 let it be equation no.2
now the equation no.2 passes through the point (-10,1)
or, 2x+5y+k =0
or, 2*-10+5*1+k= 0
or, -20+5+k= 0
or, -15+k= 0
or, k= 15
putting the value of k in equation no.2 we get,
or, 2x+5y+k=0
or, 2x+5y+15=0
which is a required line.
Answer:
there is three parts to this answer
Step-by-step explanation:
this first part is
A= 2.5
B= 49 feet
and finally
C= 0.75 seconds
Answer:
7x° + 11x° = 90°
(Required equation)
Step-by-step explanation:
7x° + 11x° = 90°
(complementary angles)
18x° = 90°
18x = 90
x = 90/18
x = 5
cos(2<em>θ</em>) + sin²(<em>θ</em>) = 0
Half-angle identity:
cos(2<em>θ</em>) + (1 - cos(2<em>θ</em>))/2 = 0
Simplify:
2 cos(2<em>θ</em>) + 1 - cos(2<em>θ</em>) = 0
cos(2<em>θ</em>) = -1
Solve for <em>θ</em> :
2<em>θ</em> = arccos(-1) + 2<em>nπ</em>
2<em>θ</em> = <em>π</em> + 2<em>nπ</em>
<em>θ</em> = <em>π</em>/2 + <em>nπ</em>
where <em>n</em> is any integer.
