Answer:
y=x\\
x=y^{2} + 12y\\
y^{2} + 12y -x = 0\\
Delta = (12^{2}) - 4.1.(-x) = 144 +4X = 36.(4+x)
\sqrt{Delta} = 6 . \sqrt{(4+x)} \\
y' = \frac{-12 + 6.(\sqrt{(4+x)}}{2} = -6 + 3.\sqrt{(4+x)}\\
y" = -6 - 3.\sqrt{(4+x)}\\\\
y' = 3.\sqrt{(4+x)} - 6\\
y''= -3.\sqrt{(4+x)} - 6\\
Answer:
The needed line equation is : x + y + 2 = 0
Step-by-step explanation:
The slope m of the given line equation = -1
The points through which the line passes is (1,-3).
Now,as we know:
The POINT SLOPE form of the equation with point (x0,y0) is given as:
y - y0 = m (x-x0)
So, the equation if line q=with m = -1 and (x0,y0) = (1,-3) is given as
y - (-3) = -1( x - 1)
or, y + 3 = 1 -x
or, x + y + 2 = 0
Hence, the needed line equation is : x + y + 2 = 0
Answer:
22.3 ft
Step-by-step explanation:
we know the base is 36 ft and we want to know the height. We know the angle of elevation
AC is you picture is 36 and we are looking for BC
We know angle A is 25
We will use tangent
tan A = opposite side/ adjacent side
tan A = BC/AC
tan 25 = BC/36
Multiply each side by 36
36 tan 25 = BC
16.78707569 = BC
But we know that he is 5.5 ft tall so the flagpole is 5.5 ft taller than BC
The triangle is from his eyes to the pole
Add 5.5 to BC to find
the height of the flagpole is BC +5.5
16.787 +5.5
22.28707569
To the nearest tenth
The flagpole is 22.3 ft
