Let’s consider the curve: r(t) = t²i +(int)j + 1/t k
X = t², y = int,z = 1/t
Utilizing x = t² and z = 1/t
X = (1/z)²
Xz² = 1
Now using y = int and z= 1/t
Y = in│1/z│
By using x = t² and y = int
Y = int = in(√x)
Thus, the resulting surfaces are,
Xz² = 1
Y = in│1/z│
Y= in(√x)
1/2x + y = 3
Rearrange the equation to achieve the slope-intercept form. (y=mx+b)
1/2x + y = 3
Eliminate 1/2x from both sides.
y = -1/2x + 3
The y-intercept (where the line intersects the y-axis) is three, and the slope (the change in y over the change in x) is -1/2.
Hope this helps!
Greetings:
<span>x² + y² + 8x + 22y + 37 = 0
(x² +8x) +(y² +22y) +37 = 0
</span>(x² +8x+4²)-4² +(y² +22y+11²) -11²+37 = 0
(x+4)² +(y+11)²-16-121+37 =0
(x+4)² +(y+11)² =10²...(<span>standard form )
</span><span>The circle's center is located at (-4, -11) and has a radius of 10</span>
It is given that K represents the midpoint of line segment JL, and KL measures 14. We must demonstrate that JK is also 14. Since K is established as the midpoint of JL and KL equals 14, it follows from the definition of midpoint that JK must also equal KL. Hence, JK is found to be 14 due to the transitive property.
