Answer for #12 Is Length = 630FT & Width 335 FT
Answer for #13 is 4600 Items & 7800 Items. I have this workbook with the same problems :) Hope this helped :)
These are just a few of the things you will learn in 6th grade. You will learn how to write a two- variable equation, how to identify the graph of an equation, graphing two-variable equations. how to interpret a graph and a word problem, and how to write an equation from a graph using a table, two-dimensional figures,Identify and classify polygons, Measure and classify angles,Estimate angle measurements, Classify triangles, Identify trapezoids, Classify quadrilaterals, Graph triangles and quadrilaterals, Find missing angles in triangles, and a lot more subjects. <span><span><span>Find missing angles in quadrilaterals
</span><span>Sums of angles in polygons
</span><span>Lines, line segments, and rays
</span><span>Name angles
</span><span>Complementary and supplementary angles
</span><span>Transversal of parallel lines
</span><span>Find lengths and measures of bisected line segments and angles
</span><span>Parts of a circle
</span><span>Central angles of circles</span></span>Symmetry and transformations
<span><span>Symmetry
</span><span>Reflection, rotation, and translation
</span><span>Translations: graph the image
</span><span>Reflections: graph the image
</span><span>Rotations: graph the image
</span><span>Similar and congruent figures
</span><span>Find side lengths of similar figures</span></span>Three-dimensional figures
<span><span>Identify polyhedra
</span><span>Which figure is being described
</span><span>Nets of three-dimensional figures
</span><span>Front, side, and top view</span></span>Geometric measurement
<span><span>Perimeter
</span><span>Area of rectangles and squares
</span><span>Area of triangles
</span><span>Area of parallelograms and trapezoids
</span><span>Area of quadrilaterals
</span><span>Area of compound figures
</span><span>Area between two rectangles
</span><span>Area between two triangles
</span><span>Rectangles: relationship between perimeter and area
</span><span>compare area and perimeter of two figures
</span><span>Circles: calculate area, circumference, radius, and diameter
</span><span>Circles: word problems
</span><span>Area between two circles
</span><span>Volume of cubes and rectangular prisms
</span><span>Surface area of cubes and rectangular prisms
</span><span>Volume and surface area of triangular prisms
</span><span>Volume and surface area of cylinders
</span><span>Relate volume and surface area
</span><span>Semicircles: calculate area, perimeter, radius, and diameter
</span><span>Quarter circles: calculate area, perimeter, and radius
</span><span>Area of compound figures with triangles, semicircles, and quarter circles</span></span>Data and graphs
<span><span>Interpret pictographs
</span><span>Create pictographs
</span><span>Interpret line plots
</span><span>Create line plots
</span><span>Create and interpret line plots with fractions
</span><span>Create frequency tables
</span><span>Interpret bar graphs
</span><span>Create bar graphs
</span><span>Interpret double bar graphs</span><span>
</span></span><span>
</span></span>
B. I hope this help please let me know if it's correct.
The word "associative" comes from "associate" or "group";the Associative Property is the rule that refers to grouping. For addition, the rule is "<span>a + (b + c) = (a + b) + c</span><span>"; in numbers, this means
</span>2 + (3 + 4) = (2 + 3) + 4. For multiplication, the rule is "<span>a(bc) = (ab)c</span>"; in numbers, this means2(3×4) = (2×3)4<span>. Any time they refer to the Associative Property, they want you to regroup things; any time a computation depends on things being regrouped, they want you to say that the computation uses the Associative Property.</span>
Answer: y = 6 mi. .
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Explanation:
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Area of a triangle = (½) * (base) * (height) ;
or, A = (½) * b * h ; or, A = b*h / 2 ;
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Given: A = 24.3 mi ² ;
b = 8.1 mi
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Find the height, "h" ; (in units of "miles", or , "mi" ).
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Plug in the known values into the formula:
24.3 mi ² = (½) * (8.1 mi) *(h) ;
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Solve for "h" (height) ;
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(½) * (8.1 mi) = 4.05 mi ;
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Rewrite:
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24.3 mi² = (4.05 mi) *(h) ; Solve for "h" ;
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Divide each side of the equation by "(4.05 mi)" ; to isolate "h" on one side of the equation ; and to solve for "h" ;
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24.3 mi² / 4.05 mi = (4.05 mi) *(h) / 4.05 mi ;
→ 6 mi = h ; ↔ h = 6 mi.
→ h = y = 6 mi.
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