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:
Im certain The answer is -4
Let me know if im wrong :)
How you would solve this problem is:
60 divided by 5 = 12
So there will be 12 toppings.
4g/cm^3
D = M/V
100g/25cm^3 = 4g/cm^3
Answer:
Well first do you have a graph or what
Step-by-step explanation:
