I need some help.. I've already done these problems and i got them back and they need fixed.. will someone please help me

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I need some help.. I've already done these problems and i got them back and they need fixed.. will someone please help me

Mathematics
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find the surface area of the solids. the pyramids are regular and the prisms, cylinders, and cones are right
For the first one, you can find the surface area of the cylinder part ( the part that looks like a can) by using 2(pi)r * height
Then add the area of the bottom circle (pi)r^2

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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
For the 2nd picture, there are 6 sides that are rectangles, so that would be 6(base*height). I can't tell in the picture, is the height of each triangle 5, or what does the 5 go to?
to the slant eight of the triangle
height*
6 there are 6 triangles, each with a base of 3 and slant height of 5.....that would be 6 * 1/2 (base * 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
1 Attachment
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

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