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Will appreciate if a mathematics guru can help out
1;The floor of a classroom is 12.8m by 10.39m, Calculate correct to 1 decimal place the perimeter of the floor.

2: Find the sum of the first perfect two square, 5, 3, 7, 4

#1. 46.4m.
for #2 i don't seem to understand the question.
not a math guru anyway, just love trying out sums

1) perimeter of a rectangular floor= 2(length + breadth). = 2(12.8 + 10.39) = 2(23.19) = 46.38 => 46.4 (1 d.p)

Question 2 is nt explicit enough. Let me use my discretion. The sum of two perfect squares i can make out from 5, 3, 7, 4, either by adding two or more figures together. It implies, 5 + 4 = 9( a perfect square of 3), and 5 + 4 + 7 = 16( a perfect square of 4), => 9 + 16 = 25. But then, figure '4' among those figures is a perfect square. But there is no other perfect square among the figures to sum up with 4, thus my take or reason for adding up figures to get two perfect squares from the figures given.

matrix199:
1) perimeter of a rectangular floor= 2(length + breadth).
= 2(12.8 + 10.39) = 2(23.19) = 46.38 => 46.4 (1 d.p)

Thanks, you nailed it and liked it

McDonie:
#1. 46.4m.

for #2 i don't seem to understand the question.

not a math guru anyway, just love trying out sums

Thanks, you nailed it and liked it

The first two perfect squares are 1 and 4.
1+4=5

matrix199:
Question 2 is nt explicit enough. Let me use my discretion. The sum of two perfect squares i can make out from 5, 3, 7, 4, either by adding two or more figures together. It implies, 5 + 4 = 9( a perfect square of 3), and 5 + 4 + 7 = 16( a perfect square of 4), => 9 + 16 = 25. But then, figure '4' among those figures is a perfect square. But there is no other perfect square among the figures to sum up with 4, thus my take or reason for adding up figures to get two perfect squares from the figures given.

1 is also a perfect square