8th Maths Practice Paper-1 1. An altitude of a triangle is 5/3 of its corresponding base. If the altitude were increased by 4cm and the base decreased by 2cm‚ the area of the triangle would remain the same. Find the base and altitude of the triangle. 2. Some toffees are bought at the rate of 11 for Rs10 and same numbers are bought at the rate of 9 for Rs 10. If the whole lot is sold at one rupee per toffee‚ find the gain or loss percent. 3. Chandu purchased a watch at 20% discount
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In Parts A and C‚ the relationship between surface area and volume was investigated. Plasticine was formed into a cube and a sphere; both shapes were cut in half. It was found that plasticine volume should not vary‚ two halves have a greater surface area than a whole‚ and cubes have a greater surface area than spheres of the same volume. In Part B‚ the relationship between diffusion and surface area to volume ratio was investigated. Three agar-phenolphthalein-sodium hydroxide cubes of different sizes
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Performance Task in GEOMETRY * Computation of the surface area‚ amount and type of needed material and the volume of the package. Volume V= L x H x W = (23 cm) (4 cm) (12cm) = (276) (4) = 1 104 cm Area A= L x W = (23cm) (12cm) = 276cm Surface Area A= 2(Lh) + 2(Lw) + 2(Wh) / 2( lh + lw + wh) = 2(23*4) + 2(23*12) + 2(12*4) = 2(92) + 2(276) + 2(48)
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Surface Area to Volume Ratio and the Relation to the Rate of Diffusion Aim and Background This is an experiment to examine how the Surface Area / Volume Ratio affects the rate of diffusion and how this relates to the size and shape of living organisms. The surface area to volume ratio in living organisms is very important. Nutrients and oxygen need to diffuse through the cell membrane and into the cells. Most cells are no longer than 1mm in diameter because small cells enable nutrients and oxygen
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[Type text] [Type text] [Type text] _An experiment on the effect of surface area to volume ratio on the rate of osmosis of Solanum tuberosum L._ BACKGROUND A cell needs to perform diffusion in order to survive. Substances‚ including water‚ ions‚ and molecules that are required for cellular activities‚ can enter and leave cells by a passive process such as diffusion. Diffusion is random movement of molecules in a net direction from a region of higher concentration to a region of lower concentration
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3 potato cubes using the dimensions from above. 2. Obtain TEACHER’S SIGN-OFF BEFORE you proceed! 3. Submerge the potato cubes (cells) into the iodine solution for 20 minutes. 4. While you are waiting for 10 minutes‚ calculate the surface area‚ volume‚ and surface area to volume ratio for each of the 3 potato cubes (cells). AFTER 10 Minutes 5. Carefully remove the potato cubes (cells) from the iodine solution and place them on absorbent paper towels. 6. Cut each cell in half and observe the
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Equipment 5 Risk Assessment 5 Method 6 Results 6 Analysis/Discussion 7 Conclusion 8 Acknowledgments/Bibliography 8 Table of Contents Background Research As an object becomes bigger its surface area and volume increases but the surface area to volume ratio decreases‚ this is because volume increases quicker than the surface area; as volume is three dimensional. This concept applies to cells and reaction rate because cells need to absorb their food‚ water and oxygen through their cell membrane and depending
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10.6 SURFACES IN SPACE EXAMPLE 6.1 Sketching a Surface © The McGraw-Hill Companies‚ Inc. Permission required for reproduction or display. Slide 1 10.6 SURFACES IN SPACE EXAMPLE 6.1 Sketching a Surface Solution Since there are no x’s in the equation‚ the trace of the graph in the plane x = k is the same for every k. This is then a cylinder whose trace in every plane parallel to the yz-plane is the parabola z = y2. © The McGraw-Hill Companies‚ Inc. Permission required for
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To investigate the relationship between surface area : volume ratio and heat loss. INTRODUCTION: The aim of this experiment is to investigate and find the relationship between heat loss (of water) and surface area to volume ratio of animals. To investigate this‚ we are going to use three flasks of different volume (as the equivalent the animals) and thus different surface areas filled with water. BACKGROUND: Surface Area : Volume Ratios We will be using the following formula for calculating
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Surface Tension of Liquids Karen Mae L. Fernan Department of Chemistry‚ Xavier University-Ateneo de Cagayan‚ Philippines Date performed: Nov. 22‚ 2012 ∙ Date Submitted: January 16‚ 2013 E-mail: fernankarenmae26@yahoo.com ------------------------------------------------------------------------------------------------------------------------------- Abstract Surface tension is defined as the energy or work required to increase the surface area of a liquid due to intermolecular forces
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