305 to the nearest ten is 310.
Yes 2,1 is a solution of this problem
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
a. a=40° (alternate interior angles)
b+40°=180°( supplementary angles)
b=140°
c= d= 140°( vertically opposite angles)
d=b=140°( corresponding angles)
b. 2a+120°=180° (supplementary angles)
2a=60°
a=30°
c. a+110°=180°( supplementary angles)
a=70°
b+70°=180°( supplementary angles)
b=110°
c+110°=180°( supplementary angles)
c=70°
c=d=70( corresponding angles)