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:x= -54.7
Step-by-step explanation:
-66.5-66.5= -133+ 78.3= -54.7
A <span>separable differential equation</span> is a first-order differential equation in which the expression for dy/dx can be factored as a function of x times a function of y,
that is, dy/dx = g(x) f(y). We can solve this equation by integrating both sides of the equation dy/f(y) = g(x)dx.
The answer should be 91 because for your total perimeter you add all the sides. So you already have your length of 75 which counts as 2 sides and equals up to 150. after you subtract your two sides from your perimeter you get 182. you then divide 182 by 2 to get your two remaining sides which gives you 91.
<h3>
</h3><h3>The area of rhombus PQRS is 120 m. </h3>
Step-by-step explanation:
<h3>correct me if I'm wrong.</h3>
