In order to solve for parallel, perpendicular, or neither, you have to look at the slope.
If the slope is the same for both equations, it is most likely parallel.
If it's the reciprocal (Where you flip the number and add change the signs. For example, the reciprocal of 1/2 is -2)
If the slope is not the same or the reciprocal, then it is neither.
So for the first equation, your slope is:
3x+2y=6
2y=-3x+6
y=-3/2x+3 The equation y=mx+b can help you here, where m is the slope.
Your slope is -3/2
For the second equation, your slope is -3/2 since y=-3/2x+5 is already in y=mx+b form and m is the slope.
Since both slopes are -3/2, then you have parallel equations!
(Be careful though, sometimes it will have the same slope but there will also be the same y-intercept. If that happens, it's no longer parallel, but it's the same equation. Such as y=-3/2x+1 and y=-3/2x+1. In this case there will be infinite solutions, but parallel equations have no solutions.)
I hope this helps!! Please ask if you have more questions!
Answer: h=(D²-s)/3
Step-by-step explanation: D² =3h+s
Subtract a from both sides of the equation
D²-s =3h+s-s
Simplifying,
D²-s =3h
Divide both sides by 3( to isolate 'h')
(D²-s)/3 =3h/3
Simplifying,
(D²-s)/3 =h
Hence, h= (D²-s)/3
Hope it helps!
Answer:
12 people
Step-by-step explanation:
9 divided by 3/4= 12
F(1) = 160 is given to us. We'll use it to find f(2)
f(n+1) = -2*f(n)
f(1+1) = -2*f(1) ... replace every n with 1
f(1+1) = -2*160 ... replace f(1) with 160
f(2) = -320
Now use f(2) to find f(3)
f(n+1) = -2*f(n)
f(2+1) = -2*f(2) ... replace every n with 2
f(3) = -2*(-320) ... replace f(2) with -320
f(3) = 640
Finally, use f(3) to find f(4)
f(n+1) = -2*f(n)
f(3+1) = -2*f(3) ... replace every n with 3
f(4) = -2*640 ... replace f(3) with 640
f(4) = -1280
Final Answer: -1280
Step-by-step explanation:
<h2>
<em>y = ± 5⁄2 x</em></h2><h2>
<em>y = ± 5⁄2 xThe equation is the left half of the hyperbola. The domain is ( – ∞, – 1⁄5 ]. The range is ( – ∞, ∞ ). The vertical line test indicates that this is not the graph of a function.</em></h2>
