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:
<em>(c). </em>
<em> </em>
Step-by-step explanation:
The <u><em>cosine</em></u> is the <em>"x" coordinate</em> of the terminal point on the unit circle.
Thus the answer is <em>(c). </em>
You need to distribute the numerical factors for each term in the parenthesis:

This equals

<span>The additive inverse of −5 is +5, so 18 − (−5) = 23. </span>