Answer: (-31.09 mi, -20.97 mi)
Step-by-step explanation:
If the noon is our initial time, then we have that 2:30 pm is t = 2.5 hours.
Now, the position (x, y) of the ship can be writen as:
(velocity*time*cos(angle), velocity*time*sin(angle))
or:
(15mph*t¨*cos(214°), 15mph*t*sin(214°))
then the position at t = 2.5 hours is:
(15mph*2.5h*cos(214°), 15mph*2.5h*sin(214°))
(-31.09 mi, -20.97 mi)
The roots of the polynomial <span><span>x^3 </span>− 2<span>x^2 </span>− 4x + 2</span> are:
<span><span>x1 </span>= 0.42801</span>
<span><span>x2 </span>= −1.51414</span>
<span><span>x3 </span>= 3.08613</span>
x1 and x2 are in the desired interval [-2, 2]
f'(x) = 3x^2 - 4x - 4
so we have:
3x^2 - 4x - 4 = 0
<span>x = ( 4 +- </span><span>√(16 + 48) </span>)/6
x_1 = -4/6 = -0.66
x_ 2 = 2
According to Rolle's theorem, we have one point in between:
x1 = 0.42801 and x2 = −1.51414
where f'(x) = 0, and that is <span>x_1 = -0.66</span>
so we see that Rolle's theorem holds in our function.
The answer is 22.94 because of reasons and math and decimals and stuffy I'm not going to go in to detail somebody else can because I'm not with the math stuff
Alrighty
squaer base so length=width, nice
v=lwh
but in this case, l=w, so replace l with w
V=w²h
and volume is 32000
32000=w²h
the amount of materials is the surface area
note that there is no top
so
SA=LW+2H(L+W)
L=W so
SA=W²+2H(2W)
SA=W²+4HW
alrighty
we gots
SA=W²+4HW and
32000=W²H
we want to minimize the square foottage
get rid of one of the variables
32000=W²H
solve for H
32000/W²=H
subsitute
SA=W²+4WH
SA=W²+4W(32000/W²)
SA=W²+128000/W
take derivitive to find the minimum
dSA/dW=2W-128000/W²
where does it equal 0?
0=2W-1280000/W²
128000/W²=2W
128000=2W³
64000=W³
40=W
so sub back
32000/W²=H
32000/(40)²=H
32000/(1600)=H
20=H
the box is 20cm height and the width and length are 40cm
Y=4 (6^x)
comment if you want me to explain it
