Algebra 1 Item No 48
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A square cardboard has a 2 inch square cut off from each corner. It is then folded up to create a container that holds 288 cubic inches. To find the perimeter of the original square cardboard, the problem sets up an equation relating the side length x to the volume formula of length x - 4, width x - 4, and height 2 inches. Solving the equation yields a side length of 16 inches. The perimeter is then calculated to be 64 inches.
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The Pythagorean theorem states that for any right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. The document provides an example proof of the theorem using geometric shapes and explains how to use the theorem to solve for unknown sides of right triangles. It also gives examples of using the Pythagorean theorem and taking square roots to solve for variables.
3.7lecture
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3. By setting up the volume and surface area equations and substituting one for the other, an equation for cost is derived as a function of just the radius.
4. Taking the derivative and setting it equal to zero finds the relative minimum, giving an optimal radius of 1.744cm, and substituting back gives the corresponding optimal height of 10.464cm.
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This document discusses various algorithms and problems including:
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This document discusses methods for solving pairs of linear equations in two variables: substitution, elimination, and cross-multiplication. The substitution method involves solving one equation for one variable and substituting it into the other equation. The elimination method involves multiplying the equations by constants to make coefficients equal and then adding or subtracting the equations to eliminate one variable. The cross-multiplication method involves cross-multiplying the coefficients of the equations to derive an equation with one variable that can then be solved.
MWA 10 7.1 Pythagorean
The document explains the Pythagorean theorem and provides examples of its use. It defines key terms like right triangle, hypotenuse, and how angles and sides are labeled. The theorem states that for a right triangle with sides a, b, c, where c is the hypotenuse, a^2 + b^2 = c^2. Examples show setting up and solving equations using the theorem to find missing side lengths of right triangles. The final example calculates the length of cloth needed to make a tent with a 4m high, 3m wide opening.
4.11.2 Special Right Triangles
1) Special right triangles have properties that allow for determining side lengths without using the Pythagorean theorem. The 45-45-90 triangle has legs that are equal in length and a hypotenuse that is √2 times the leg length. The 30-60-90 triangle has a shorter leg that is half the hypotenuse, a longer leg that is √3 times the shorter leg, and a hypotenuse that is 2 times the shorter leg.
2) Examples demonstrate using the properties of special right triangles to find missing side lengths by setting up and solving equations based on the corresponding theorems. Rationalizing denominators may be required when finding a leg from a known hypotenuse.
3)
Obj. 24 Special Right Triangles
Identify when a triangle is a 45-45-90 or 30-60-90 triangle
Use special right triangle relationships to solve problems
Ch04 sect05
1) Four examples of optimization problems are presented with the goal of minimizing or maximizing an objective function subject to constraints.
2) The problems are solved by taking derivatives of the objective function and setting them equal to zero to find critical points, then using the second derivative test to verify if these points give a minimum or maximum.
3) Analytical solutions are provided for each example problem, such as finding the dimensions of a box that minimize material usage given a volume constraint.
Obj. 24 Special Right Triangles
Identify when a triangle is a 45-45-90 or 30-60-90 triangle
Use special right triangle relationships to solve problems
4.11.2 Special Right Triangles
This document discusses special right triangles and their properties. It defines 45-45-90 and 30-60-90 triangles, and provides the key relationships between their sides: for 45-45-90 triangles, the hypotenuse is √2 times the leg; for 30-60-90 triangles, the hypotenuse is 2 times the shorter leg and √3 times the longer leg. It provides examples of using these relationships to solve for missing side lengths.
Geometry 1st Edition Kindle Edition by Elayn Martin Gay Solutions Manual
link full download: https://bit.ly/2GS0vdq
Language: English
ISBN-10: 0134173651
ISBN-13: 978-0134173658
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5 2nd degree equations and the quadratic formula
The document describes three methods for solving second degree equations of the form ax2 + bx + c = 0:
1) The square-root method is used when the x-term is missing. It involves solving for x2 and taking the square root to find x.
2) Factoring involves factoring the equation into the form (ax + b)(cx + d) = 0 and setting each binomial equal to 0 to solve for x.
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Similar to Algebra 1 Item No 48 (20)
Pair of linear equations in two variables for classX
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Geometry 1st Edition Kindle Edition by Elayn Martin Gay Solutions Manual
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Tags: Information Security, ISO/IEC 27001, ISO/IEC 42001, Artificial Intelligence, GDPR
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Syllabus
Chapter-1
Introduction to objective, scope and outcome the subject
Chapter 2
Introduction: Scope and Specialization of Civil Engineering, Role of civil Engineer in Society, Impact of infrastructural development on economy of country.
Chapter 3
Surveying: Object Principles & Types of Surveying; Site Plans, Plans & Maps; Scales & Unit of different Measurements.
Linear Measurements: Instruments used. Linear Measurement by Tape, Ranging out Survey Lines and overcoming Obstructions; Measurements on sloping ground; Tape corrections, conventional symbols. Angular Measurements: Instruments used; Introduction to Compass Surveying, Bearings and Longitude & Latitude of a Line, Introduction to total station.
Levelling: Instrument used Object of levelling, Methods of levelling in brief, and Contour maps.
Chapter 4
Buildings: Selection of site for Buildings, Layout of Building Plan, Types of buildings, Plinth area, carpet area, floor space index, Introduction to building byelaws, concept of sun light & ventilation. Components of Buildings & their functions, Basic concept of R.C.C., Introduction to types of foundation
Chapter 5
Transportation: Introduction to Transportation Engineering; Traffic and Road Safety: Types and Characteristics of Various Modes of Transportation; Various Road Traffic Signs, Causes of Accidents and Road Safety Measures.
Chapter 6
Environmental Engineering: Environmental Pollution, Environmental Acts and Regulations, Functional Concepts of Ecology, Basics of Species, Biodiversity, Ecosystem, Hydrological Cycle; Chemical Cycles: Carbon, Nitrogen & Phosphorus; Energy Flow in Ecosystems.
Water Pollution: Water Quality standards, Introduction to Treatment & Disposal of Waste Water. Reuse and Saving of Water, Rain Water Harvesting. Solid Waste Management: Classification of Solid Waste, Collection, Transportation and Disposal of Solid. Recycling of Solid Waste: Energy Recovery, Sanitary Landfill, On-Site Sanitation. Air & Noise Pollution: Primary and Secondary air pollutants, Harmful effects of Air Pollution, Control of Air Pollution. . Noise Pollution Harmful Effects of noise pollution, control of noise pollution, Global warming & Climate Change, Ozone depletion, Greenhouse effect
Text Books:
1. Palancharmy, Basic Civil Engineering, McGraw Hill publishers.
2. Satheesh Gopi, Basic Civil Engineering, Pearson Publishers.
3. Ketki Rangwala Dalal, Essentials of Civil Engineering, Charotar Publishing House.
4. BCP, Surveying volume 1
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This Dissertation explores the particular circumstances of Mirzapur, a region located in the
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changes, conversion trends, and other related patterns. The spatial dimensions of land use and
cover support policymakers and scientists in making well-informed decisions, as alterations in
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Algebra 1 Item No 48
- 1. Algebra 1 Item No. 48 A square cardboard (see diagram) is designed to form a container. A 2 inch square is cut off on each corner. The container can hold 288 cubic inches. Find the perimeter of the square. A.8 in C. 32 in B. 16 in D. 64 in www.upcatreview.com 2 in. 2 in. 2 in.
- 2. 2 in. 2 in. 2 in. Volume of container = 288 cm3 To solve the problem, we must first visualize the given situation: www.upcatreview.com
- 3. We then make a representation of the unknown: Let x be the length of the side of the cardboard. 2 in. 2 in. 2 in. x x – 4 By subtracting 2 inches from both ends of the side, we get the representation for the box’s length and width, x – 4 www.upcatreview.com
- 4. The cardboard will be folded and will then be made into a box like so: 2 in. Volume of container = 288 cm3 This will be the height of the container. This will be the length (and also the width) of the container. www.upcatreview.com x – 4
- 5. We will use the formula in finding the volume: V = LWH www.upcatreview.com
- 6. We will use the formula in finding the volume: V = LWH The length L can be expressed in terms of x: L = x – 2 – 2 or L = x – 4 www.upcatreview.com
- 7. We will use the formula in finding the volume: V = LWH The length L can be expressed in terms of x: L = x – 2 – 2 or L = x – 4 Same is true for the width W: W = x – 2 – 2 or W = x – 4 www.upcatreview.com
- 8. We will use the formula in finding the volume: V = LWH The length L can be expressed in terms of x: L = x – 2 – 2 or L = x – 4 Same is true for the width W: W = x – 2 – 2 or W = x – 4 The height H of the container would be 2 inches. www.upcatreview.com
- 9. Putting everything in its proper place in the equation, we have: 2 )4(144 2)4)(4(288 −= −−= = x xx LWHV www.upcatreview.com
- 10. Putting everything in its proper place in the equation, we have: Solving the quadratic equation by getting the square root of both sides, we get: 2 )4(144 2)4)(4(288 −= −−= = x xx LWHV 8or16 124 4144 −== =± −=± xx x x We reject the negative solution. www.upcatreview.com
- 11. Note that x = 16 is NOT yet the final answer. We should always remember that we have to check whether we have answered the real question in the item. www.upcatreview.com
- 12. Note that x = 16 is NOT yet the final answer. We should always remember that we have to check whether we have answered the real question in the item. We are asked for the perimeter. To obtain it: in64 )16(4 4 = = = sPerimeter www.upcatreview.com
- 13. The answer is: D. 64 in Algebra 1 Item No. 48 www.upcatreview.com
- 14. Algebra 1 Item No. 48 A square cardboard (see diagram) is designed to form a container. A 2 inch square is cut off on each corner. The container can hold 288 cubic inches. Find the perimeter of the square. A.8 in C. 32 in B. 16 in D. 64 in 2 in. 2 in. 2 in. www.upcatreview.com