The length of a rectangle is 6.7cm correct to 2.s.f.
(i) the upper bound
(ii) lower bound
The area of the rectangle is 26.9cm2 correct to 3 s.f.
(i) calculate the upper bound for the width of teh rectangle write down all the figures on calculator.
(ii) Calculate the lower bound for the width of the rectangle.
Please try and exaplain every step in this question, as i have a maths GCSE retake next week so please help!!
Thank you!
2006-10-30 18:11:44 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
The length can be between 6.68 cm (lower bound) and 6.72 cm (upper bound), correct to 2 s.f..
The area can be between 26.87 and 26.93 cm2, correct to 3 s.f.
(i) Taking upper bound area of 26.93 cm2 and length of upper bound as 6.72 cm, its width would be 26.93/6.72, the width will be 4.0074404 or rounded it is 4 cm.
(ii) Taking lower bound area of 26.87 and length of lower bound as 6.68 cm, its width would be 26.87 cm2 / 6.68 cm = 4.022455 cm. which when rounded also comes to 4 cm.
Hope this helps !!
2006-10-30 20:31:43 · answer #1 · answered by young_friend 5 · 0⤊ 1⤋
The length can be between 6.65 and 6.75 cm.
All the possible values are rounded to 6.7 cm
The area can be between 26.85 and 26.95 cm2.
All the possible values are rounded to 26.9 cm2.
The max width is 26.95 cm2 / 6.65 cm = 4.05 cm.
The min. width is 26.85 cm2 / 6.75 cm = 3.98 cm.
The most probable width is (4.05 + 3.98)/2 cm = 4.0 cm or
26.9 cm2 / 6.7 cm = 4.0 cm
Th
2006-10-30 18:23:47 · answer #2 · answered by Thermo 6 · 4⤊ 0⤋
1. 80
2016-05-22 14:36:43 · answer #3 · answered by Anonymous · 0⤊ 0⤋