Mark made a poster for his room by enlarging a photo of his favourite hockey player, which he took himself. The original photo had the dimensions 6 by 9cm. Mark enlarged the picture by 180% but found it still wasn't big enough. He then had the enlarged photo enlarged again by 190%. What are the dimensions of his final poster?
Please don't just give me the answer, but how you got it as well. I've tried many times but I just can't get it right.
2006-10-25 10:36:57 · 2 answers · asked by Rei-chan 3 in Education & Reference ➔ Homework Help
ok , the initial dimensions of the photo are 6 cm by 9 cm.
To obtain the size after 180 % enlargement , you take 180% of 6 cm and add it to 6 cm and same with 9 cm.
180% of 6 cm is (180/100) * 6 cm thats (1.8)*6cm = 10.8cm and 180% of 9 cm is (1.9)*9cm = 16.2 cm.
you get the 180% enlargement + original size as:
(10.8 + 6) cm = 16.8 cm by (16.2 + 9) cm = 25.2 cm
the dimensions now of the enlarged poster are 16.8cm by 25.2 cm
Again , now the already enlarged photo has to be enlarged by 190% , so you take 190 % of 16.8 cm and add 16.8 to it and same with 25.2 cm.
you get, 190% enlargement of already enlarged pic:
(31.92 + 16.8)cm = 48.72 cm by (47.88 + 25.2)cm = 73.08cm
the dimensions you get now are 48.72 cm by 73.08 cm which is the solution to your problem.
or the final dimension of the poster are 0.4872 m by 0.7308 m.
I hope this helps you , good luck !
:)
2006-10-25 10:38:42 · answer #1 · answered by Aqua 4 · 0⤊ 1⤋
190% is just the original dimensions times 1.9 -- that's it. 100% is the original times 1, 50% is the original times .5, etc.
So 6cm times 1.9 = 11.4cm, and 9cm times 1.9 = 17.1cm. Those are the final dimensions.
2006-10-25 10:40:09 · answer #2 · answered by Anonymous · 0⤊ 2⤋