Question

Could you check my answer?
@jim_thompson5910

Answer 1

@jim_thompson5910

Answer 2

how are you getting 502.8 ?

Answer 3

I see now, thanks

Answer 4

What I would do is replace each '5' with '9*5'
and replace every '11' with '9*11'

we do this because each dimension is 9 times longer

then recompute the surface area of the new figure

Answer 5

keep in mind that 5^2 will turn into (9*5)^2

Answer 6

\[2 \times 22 \div 7 \times 9*5 + 2 \times 22 \div 7 \times 5 \times 9 * 11 = ?\]

Answer 7

you forgot to square the "9*5" portion

Answer 8

Oh that's was meant to be (9*5)^2

Answer 9

Yes.

Answer 10

also, you forgot to replace the second '5' with '9*5'

Answer 11

\[2 \times 22 \div 7 \times (9*5)^2 + 2 \times 22 \div 7 \times 9*5 \times 9 * 11 = ? \]

Answer 12

good

Answer 13

now compute that

Answer 14

I got a SUPER HUGE # !

Answer 15

what did you get

Answer 16

178218183.7

Answer 17

>.<

Answer 18

hmm, strange, it's not supposed to be that big

Answer 19

well anyways, I did

9*5 = 45
9*11 = 99

so that means...

2*(22/7)*45^2+2*(22/7)*45*99 = 40,731.42857 (approximately)

Answer 20

yeah we're rounding to the nearest whole number

Answer 21

check to make sure you get the same (and try to figure out what went wrong)

Answer 22

Yes, now I got that answer!

Answer 23

Thank you so much!

Answer 24

you're welcome