Answer:
= 1.281244e+274
Step-by-step explanation:
I have NO IDEA if that helps,... BUT I hope it does,... Chow,...!
Hope this helps but I think the answer could be : 157+25m=50+60m
For the equation F(x) = ax² + bx + c we have:
- maximum value if a<0
- minimum value if a>0
F(x) = -3x² + 18x + 3 ⇒ a = -3, b = 18
a < 0 ⇒ the function has a maximum value
Quadratic function has the maximum value (or minimum) at vertex of its parabola.
The maximum value is k at x=h where:
and k = F(h)
![h=\dfrac{-18}{2\cdot(-3)}=3\\\\F(3)=-3\cdot3^2+18\cdot3+3=-27+54+3=30](https://tex.z-dn.net/?f=h%3D%5Cdfrac%7B-18%7D%7B2%5Ccdot%28-3%29%7D%3D3%5C%5C%5C%5CF%283%29%3D-3%5Ccdot3%5E2%2B18%5Ccdot3%2B3%3D-27%2B54%2B3%3D30)
Therefore:
<h3>
The function has a maximum value of 30 at x = 3</h3>
Answer:
is irrational so any attempt to use 3.14... is never EXACT...
do not try to convert it ... if it asks for exact..
write 81
or 9
etc. don't put in 63.62 like answers
Step-by-step explanation:
Ya, calculus and related rates, such fun!
everything is changing with respect to t
altitude rate will be dh/dt and that is 1cm/min
dh/dt=1cm/min
area will be da/dt which is increasing at 2cm²/min
da/dt=2cm²/min
base=db/dt
alright
area=1/2bh
take dervitivie of both sides
da/dt=1/2((db/dt)(h)+(dh/dt)(b))
solve for db/dt
distribute
da/dt=1/2(db/dt)(h)+1/2(dh/dt)(b)
move
da/dt-1/2(dh/dt)(b)=1/2(db/dt)(h)
times 2 both sides
2da/dt-(dh/dt)(b)=(db/dt)(h)
divide by h
(2da/dt-(dh/dt)(b))/h=db/dt
ok
we know
height=10
area=100
so
a=1/2bh
100=1/2b10
100=5b
20=b
so
h=10
b=20
da/dt=2cm²/min
dh/dt=1cm/min
therefor
(2(2cm²/min)-(1cm/min)(20cm))/10cm=db/dt
(4cm²/min-20cm²/min)/10cm=db/dt
(-16cm²/min)/10cm=db/dt
-1.6cm/min=db/dt
the base is decreasing at 1.6cm/min