If
<em>f(x)</em> = <em>ax</em> ³ + <em>bx</em> ² - 5<em>x</em> + 9
then
<em>f '(x)</em> = 3<em>ax </em>² + 2<em>bx</em> - 5
Given that <em>f</em> (-1) = 12 and <em>f</em> '(-1) = 3, we get the system of equations
-<em>a</em> + <em>b</em> + 5 + 9 = 12
3<em>a</em> - 2<em>b</em> - 5 = 3
or
-<em>a</em> + <em>b</em> = -2
3<em>a</em> - 2<em>b</em> = 8
Multiply through the first equation by 2 and add it to the second one to eliminate <em>b</em> and solve for <em>a</em> :
2(-<em>a</em> + <em>b</em>) + (3<em>a</em> - 2<em>b</em>) = 2(-2) + 8
-2<em>a</em> + 2<em>b</em> + 3<em>a</em> - 2<em>b</em> = -4 + 8
<em>a</em> = 4
Substitute this into the first equation above to solve for <em>b</em> :
-4 + <em>b</em> = -2
<em>b</em> = 2
<span>trapezoid area = ((sum of the bases) ÷ 2) • height</span>
The answer is 3/-2 hope that helps!
Answer:
C) 360pi in2
Step-by-step explanation:
Given:
radius, r= 10in
height, h=26in
surface area of the cone, T.S.A= ?
T.S.A=πrl +πr^2
=π(10)(26) +π(10)^2
=260π+100π
=360π^2 !
