Y=mx+c
m1 x m2 = - 1 for perpendicular lines
m= -1/5
so now y= - 1/5x + c
Substitute given coordinate
3= (-1/5 × 3) +c
3= -3/5 + c
add 3/5 to both sides
2.6 (a.k.a 18/5) = y
So therefore:
y=(-x+18)/5
I think this is correct
Answer:
Step-by-step explanation:
3x2 - y2 = 9.........(1)
x2 + 2y2 = 38......(2)
Let x^2 = a and y^2 = b
:- the two equations becomes
3a - b = 9..........(3)
a + 2b = 38.......(4)
Multiplying (4) by 3
3a + 6b = 114........(5)
Substracting (3) from (5)
3a - 3a + 6b - (-b) = 114 - 9
6b +b = 105
7b = 105
b = 105/7
b = 15
And 3a - b = 9 putting b = 15
3a - 15 = 9
3a = 15+9
3a = 24
a = 24/3
a = 8
And
x^2 = a
x^2 = 8
x = √8
x = √(4×2)
x = 2√2
Also y^2 = b
y^2 = 15
y = √15
It means find the angle whose tangent equals - sqrt (3)
the negative square root of 3 =
(-<span>
<span>
<span>
1.7320508076)
</span>
Looking up the arc tan </span></span><span>(-1.7320508076</span> ) on this page:
http://www.1728.org/trigcalc.htm
we find it equals -60 degrees
F(x)=2x^2+5x-8 is an example of a non-linear equation. Anything with x is linear, anything with x raised to a power (any power except 0 and 1) is non-linear.