Answer:
The equation of the boundary of the region is (x+2)² + (y-1)² = 25
Step-by-step explanation:
The satellite is positioned at (-2,1) and can sense the presence within 5 miles of its position.
The boundary of the region is a circle with center (-2,1) and radius 5.
The equation of the circle with center (h,k) and radius r is given by
(x-h)²+(y-k)²=r²
The equation of the circle with center (-2,1) and radius 5 is given by
(x+2)²+(y-1)²=5²=25
(x+2)² + (y-1)² = 25 is the required equation of the boundary of the region.
Use the chain rule:
<em>y</em> = tan(<em>x</em> ² - 5<em>x</em> + 6)
<em>y'</em> = sec²(<em>x</em> ² - 5<em>x</em> + 6) × (<em>x</em> ² - 5<em>x</em> + 6)'
<em>y'</em> = (2<em>x</em> - 5) sec²(<em>x</em> ² - 5<em>x</em> + 6)
Perhaps more explicitly: let <em>u(x)</em> = <em>x</em> ² - 5<em>x</em> + 6, so that
<em>y(x)</em> = tan(<em>x</em> ² - 5<em>x</em> + 6) → <em>y(u(x))</em> = tan(<em>u(x)</em> )
By the chain rule,
<em>y'(x)</em> = <em>y'(u(x))</em> × <em>u'(x)</em>
and we have
<em>y(u)</em> = tan(<em>u</em>) → <em>y'(u)</em> = sec²(<em>u</em>)
<em>u(x)</em> = <em>x</em> ² - 5<em>x</em> + 6 → <em>u'(x)</em> = 2<em>x</em> - 5
Then
<em>y'(x)</em> = (2<em>x</em> - 5) sec²(<em>u</em>)
or
<em>y'(x)</em> = (2<em>x</em> - 5) sec²(<em>x</em> ² - 5<em>x</em> + 6)
as we found earlier.
Change the mixed number to an improper fraction:
=
=5 ×
=
=8
Answer is
C. 8
The answer would be true, because the 2 on the bottom of the fraction would cancel out the 2 in the equation.
