Please helpppppp!!! ASAP
1 answer:
You might be interested in
Answer:
The found values are:
y = 10
<JKM = 106°
<MKL = 74°
Step-by-step explanation:
(See the diagram attached)
As JKL is a straight horizontal line, angle measure from JK to KL is 180°.
We can see in the diagram that line KM is dividing this angle of 180° into 2 unequal parts.
Which means that the sum of <JKM and <MKL is equals to 180°.
Mathematically, it can be written as:
Substitute the values of both angle to solve the equation:
Put in the formulas of both angle to find their values:
<JKM = 10y + 6
<JKM = 10(10)+6
<JKM = 106°
<MKL = 8y - 6
<MKL = 8(10)-6
<MKL = 74°
Answer:
(a) t = ±2
(b) t ∈ {0, 1}
(c) In navigation terms: east by north. The slope is about 0.42 at that point.
Step-by-step explanation:
(a) dy/dx = 0 when dy/dt = 0
dy/dt = 3t^2 -12 = 0 = 3(t -2)(t +2)
The slope is zero at t = ±2.
__
(b) dy/dx = (dy/dt)/(dx/dt) = <em>undefined</em> when dx/dt = 0
dx/dt = 6t^2 -6t = 6(t)(t -1) = 0
The slope is undefined for t ∈ {0, 1}.
__
(c) At t=3, dy/dx = (dy/dt)/(dx/dt) = 3(3-2)(3+2)/(6(3)(3-1)) = 15/36 = 5/12
The general direction of movement is away from the origin along a line with a slope of 5/12, about 22.6° CCW from the +x direction.
_____
The first attachment shows the derivative and its zeros and asymptotes. It also shows some of the detail of the parametric curve near the origin.
The second attachment shows the parametric curve over the domain for which it is defined, along with the point where t=3.
