Qn. 1
Lower bound for Zoe's weight = 62 - (1/2) = 62 - 0.5 = 61.5 kg
Qn. 2
Upper bound for length AB = 8.3+ (0.1/2) = 8.3+0.05 = 8.35 cm
Qn. 3
Upper bound for Anu's wight = 83+(0.5/2) = 83+0.25 = 83.25 kg
Qn. 4
Lower bound for length CD = 27-(0.5/2) = 27-0.25 = 26.75 cm
Qn. 5
Upper bound for sides of the hexagon = 3.6+(0.1/2) = 3.6+0.05 = 3.65 cm
Upper bound for the perimeter = upper bound for the sides*6 = 3.65*6 = 21.9 cm
Qn. 6
Perimeter = 4*length => side = Perimeter/4 = 24/4 = 6
Bound = 0.5/4 = 0.125
Lower bound of the length = 6-0.125 = 5.875 cm
Qn. 7
For the area,
Upper bound = 80+(10/2) 80+5 = 85 cm^2
For the length
Upper bound = 12+(1/2) = 12+0.5 = 12.5
Then, upper bound for the width = Upper bound for the area/upper bound for the length = 85/12.5 = 6.8 cm
Qn. 8
Lower bound for the area = 230-(1/2) = 230-0.5 = 229.5 cm^2
Lower bound for the sides of the square = Sqrt(Lower bound of the area) = Sqrt (229.5) = 15.15
Then,
Lower bound of perimeter = 4(Length) = 4*15.15 = 60.6 cm
Answer: 
Step-by-step explanation:
Given : The height of the rectangle = 
The width of the rectangle = 
Formula : Area = height x width
Therefore , the area of triangle in terms of polynomial will be :
![6k^3\times( 2k^2+4k+5)\\\\= 6k^3(2k^2)+6k^3(4k)+6k^3(5)\ \ [\text{Using Distributive property}]\\\\=12k^{3+2}+24k^{3+1}+30k^3\ \ [\text{Using exponents rule}:\ a^n\times a^m=a^{n+m}]\\\\=12k^5+24k^4+30k^3](https://tex.z-dn.net/?f=6k%5E3%5Ctimes%28%202k%5E2%2B4k%2B5%29%5C%5C%5C%5C%3D%206k%5E3%282k%5E2%29%2B6k%5E3%284k%29%2B6k%5E3%285%29%5C%20%5C%20%5B%5Ctext%7BUsing%20Distributive%20property%7D%5D%5C%5C%5C%5C%3D12k%5E%7B3%2B2%7D%2B24k%5E%7B3%2B1%7D%2B30k%5E3%5C%20%5C%20%5B%5Ctext%7BUsing%20exponents%20rule%7D%3A%5C%20a%5En%5Ctimes%20a%5Em%3Da%5E%7Bn%2Bm%7D%5D%5C%5C%5C%5C%3D12k%5E5%2B24k%5E4%2B30k%5E3)
Hence, the area of the entire rectangle =
The answer is A. 2, -2
Hope I helped!
Let me know if you need anything else!
~ Zoe
You have to do length of the base x width of the base x height of pyramid
Can salvage 15c+ -15c is 0 + 8 is still 0
