Tan=o/adj
Tan=60/b
B=60/tan(68)
I don't have a calculator on me but you should be able to do it from there
Differentiate implicitly using the power rule:
<span>12x^2 + 15y^2y' = 0 </span>
<span>Make y' the subject: </span>
<span>15y^2y' = -12x^2 </span>
<span>y' = -(4x^2) / 3y^2 </span>
<span>Differentiate again using the quotient rule: </span>
<span>y'' = [-8x(3y^2) - 6yy'(-4x^2)] / 9y^4 </span>
<span>y'' = (-24xy^2 + 24x^2yy') / 9y^4 </span>
<span>Simplify: </span>
<span>y'' = (-8xy + 8x^2y') / 3y^3 </span>
<span>Substitute (4x^2) / 3y^2 back for y': </span>
<span>y'' = (-8xy + 8x^2 * 4x^2 / 3y^2) / 3y^3 </span>
<span>y'' = (-24xy^3 + 32x^4) / 9y^5</span>
Answer:
Step-by-step explanation:
2)f(x) = 2x² +4x - 3
a = 2 ; b = 4 ; c = -3
1) Put x = -1 in the equation
f(x) = 2*(-1)² + 4*(-1) -3 = 2 - 4 -3 = -5
Vertex = (-1,-5)
2) Upward
3) Minimum
4) axis of symmetry = -b/2a = -4/2*2=-4/4= -1
x = -1
5) domain: all real numbers (-∞ ,∞)
Range : y ≥ -5 ; [-5 , ∞)
3) f(x) = 3x² - 6x + 4
Vertex : (1,1)
Opening : upward
Minimum
Axis of symmetry: x = 1
Domain: all real numbers
Range: y ≥ 1 ; [1, ∞)
4)f(x) = -x² - 2x - 3
a) Vertex: f(x) = -(-1)² - 2*(-1) - 3 = -1 + 2 - 3 = -2
Vertex( -1,-2)
b) downward
c) Maximum
d) Axis of symmetry: x = 2/-2 = -1
x = -1
e) Domain: all real numbers
Range: y ≤ -2 ; (-∞ , -2]
5)f(x) = 2(x -2)²
a) Vertex: (2, 0)
b) Opening: upward
c) Minimum
d) x = 2
e)Domain: all real numbers
Range: y ≥ 0 ; [0,∞)
You need to put the whole question but I have a question like this and it’s 9 feet
Step-by-step explanation:
Look at the pictures.
1. From point P, we draw an arc crossing the line m at two points.
2. From points F and G, we draw intersecting arcs with the same radius.
3. Through P and L, we draw a line perpendicular to the m line.