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find the surface area of the solids. the pyramids are regular and the prisms, cylinders, and cones are right

Then add the area of the bottom circle (pi)r^2

and then add SA of the cone, pi*r*s
s is the slant height

thats what i did and i got 211.95cmin^2

ok, let me try and work it out with the numbers, just a sec

Do you know the right answer?

no

I got 203.47cm^2

wekkm 203.58 cm^2 when I use the pi button

oops, I mean "well"

for the cylinder, it's 30.6pi, the circle base is 9pi, and the cone is 25.2pi

so it would be 30.6pi+9pi+25.2pi

yes, and with that I got 64.8pi

to the slant eight of the triangle

height*

So, 6(3*6) +6(1/2)(3*5)

and then would you have to add the area of the base?

855+64.8pi

where did you get 855?

6(3*6) +6(1/2)(3*5)

6(3*12) +6(1/2)(3*5)
that would be a 12 in the first ( )

153

64.8pi went to the first drawing...the one above is for the 2nd picture

I get 261 in^2

i thought the first pic was 203.58cm^2

yes, that's what I got....it was 64.8pi, which = 203.58cm^2

ok

for the 2nd drawing I get 261in^2 + area of that hexagon base

261 in^2 so this is the second pic surface area?

yes, but I dont' know how to find area of the base

will u help me with some other ones that ive already done i just got them wrong

i will try

okay.. for this one i needed to find the value of x.. and i got 3

v=1/3 pi r^2 h

I get 3 also

so redoing it i get \[56.5=1/3\times \pi \times x ^{2}\times6\]

3 meters

\[56.5=6.28x ^{2}\]

yes, i get that too

Did it count off because you didn't put meters?

maybe

oh, wait, is x the diameter? We only found the radius.

so x = 6 meters

56.5/6.28=8.996

yeah, then sq rt of that gives the radius of that circle as 3 meters

but in the drawing, the x represents the diameter, which would be twice the radius, or 6 meters

I've got to go....good luck!

thank u

you're welcome