1/4 = 3/12
3/12 < 5/12
Therefore, 5/12 is greater than 1/4
I believe it's four times as great : )
Answer:
-g + 4
Step-by-step explanation:
X + 68 with a remainder of (-592)
Answer:
<h3><u>Question 7</u></h3>
<u>Lateral Surface Area</u>
The bases of a triangular prism are the triangles.
Therefore, the Lateral Surface Area (L.A.) is the total surface area excluding the areas of the triangles (bases).
![\implies \sf L.A.=2(10 \times 6)+(3 \times 6)=138\:\:m^2](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%20L.A.%3D2%2810%20%5Ctimes%206%29%2B%283%20%5Ctimes%206%29%3D138%5C%3A%5C%3Am%5E2)
<u>Total Surface Area</u>
Area of the isosceles triangle:
![\implies \sf A=\dfrac{1}{2}\times base \times height=\dfrac{1}{2}\cdot3 \cdot \sqrt{10^2-1.5^2}=\dfrac{3\sqrt{391}}{4}\:m^2](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%20A%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20base%20%5Ctimes%20height%3D%5Cdfrac%7B1%7D%7B2%7D%5Ccdot3%20%5Ccdot%20%5Csqrt%7B10%5E2-1.5%5E2%7D%3D%5Cdfrac%7B3%5Csqrt%7B391%7D%7D%7B4%7D%5C%3Am%5E2)
Total surface area:
![\implies \sf T.A.=2\:bases+L.A.=2\left(\dfrac{3\sqrt{391}}{4}\right)+138=167.66\:\:m^2\:(2\:d.p.)](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%20T.A.%3D2%5C%3Abases%2BL.A.%3D2%5Cleft%28%5Cdfrac%7B3%5Csqrt%7B391%7D%7D%7B4%7D%5Cright%29%2B138%3D167.66%5C%3A%5C%3Am%5E2%5C%3A%282%5C%3Ad.p.%29)
<u>Volume</u>
![\sf \implies Vol.=area\:of\:base \times height=\left(\dfrac{3\sqrt{391}}{4}\right) \times 6=88.98\:\:m^3\:(2\:d.p.)](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%20Vol.%3Darea%5C%3Aof%5C%3Abase%20%5Ctimes%20height%3D%5Cleft%28%5Cdfrac%7B3%5Csqrt%7B391%7D%7D%7B4%7D%5Cright%29%20%5Ctimes%206%3D88.98%5C%3A%5C%3Am%5E3%5C%3A%282%5C%3Ad.p.%29)
<h3><u>Question 8</u></h3>
<u>Lateral Surface Area</u>
The bases of a hexagonal prism are the pentagons.
Therefore, the Lateral Surface Area (L.A.) is the total surface area excluding the areas of the pentagons (bases).
![\implies \sf L.A.=5(5 \times 6)=150\:\:cm^2](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%20L.A.%3D5%285%20%5Ctimes%206%29%3D150%5C%3A%5C%3Acm%5E2)
<u>Total Surface Area</u>
Area of a pentagon:
![\sf A=\dfrac{1}{4}\sqrt{5(5+2\sqrt{5})}a^2](https://tex.z-dn.net/?f=%5Csf%20A%3D%5Cdfrac%7B1%7D%7B4%7D%5Csqrt%7B5%285%2B2%5Csqrt%7B5%7D%29%7Da%5E2)
where a is the side length.
Therefore:
![\implies \sf A=\dfrac{1}{4}\sqrt{5(5+2\sqrt{5})}\cdot 5^2=43.01\:\:cm^2\:(2\:d.p.)](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%20A%3D%5Cdfrac%7B1%7D%7B4%7D%5Csqrt%7B5%285%2B2%5Csqrt%7B5%7D%29%7D%5Ccdot%205%5E2%3D43.01%5C%3A%5C%3Acm%5E2%5C%3A%282%5C%3Ad.p.%29)
Total surface area:
![\sf \implies T.A.=2\:bases+L.A.=2(43.01)+150=236.02\:\:cm^2\:(2\:d.p.)](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%20T.A.%3D2%5C%3Abases%2BL.A.%3D2%2843.01%29%2B150%3D236.02%5C%3A%5C%3Acm%5E2%5C%3A%282%5C%3Ad.p.%29)
<u>Volume</u>
![\sf \implies Vol.=area\:of\:base \times height=43.011193... \times 6=258.07\:\:cm^3\:(2\:d.p.)](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%20Vol.%3Darea%5C%3Aof%5C%3Abase%20%5Ctimes%20height%3D43.011193...%20%5Ctimes%206%3D258.07%5C%3A%5C%3Acm%5E3%5C%3A%282%5C%3Ad.p.%29)