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:
It's closer to 1
Step-by-step explanation:
Answer:
120
Step-by-step explanation:
Here, we want to get the possible value of A
Now, we should understand that the value of an angle in triangle cannot be more than 180
Also, cosine can only be negative in the second and the 3rd quadrant
So,
arc cos (-0.5) = 120 , 300 , 60
But 60 is not correct as it will give a negative value
300 is not correct also as it is more than the angles present in a triangle
Thus, our answer is 120
