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[3.NF.3a-1.0] Intro Fraction Equivalence - Common Core Standard
By Freckle education
Explain equivalence of fractions in special cases: Understand two fractions as equivalent (equal) if they are the same size, or the same point on a number line.
[3.NF.1-1.0] What is a fraction - Common Core Standard
By Freckle education
Understand a fraction 1/b as the quantity formed by 1 part when a whole is partitioned into b equal parts
[3.NF.1-1.0] What is a fraction - Common Core Standard
By Front Row
Understand a fraction 1/b as the quantity formed by 1 part when a whole is partitioned into b equal parts
Equivalent fractions using the number line (3.NF.3A)
By Far East Software
Understand two fractions as equivalent (equal) if they are the same size, or the same point on a number line.
Understanding fractions (3.NF.1)
By Far East Software
Understand a fraction 1/b as the quantity formed by 1 part when a whole is partitioned into b equal parts; understand a fraction a/b as the quantity formed by a parts of size 1/b.
Quadratic Word Problems | MathHelp.com
By MathHelp.com
A number is 56 less than its square. Find the number. To solve this problem, let’s translate the first sentence into an equation. A number, that’s x, is, =, 56 less than it’s square, that’s x squared – 56. Remember that “less than” switches the order around. In other words, “56 less than its square” is not 56 minus x squared, it’s x squared minus 56. Next, since we have an x squared term in our equation, we set it equal to 0 by subtracting x from both sides, and we have 0 = x squared – x – 56. Next, we factor the right side as the product of two binomials. In the first position of each binomial, we have the factors of x squared, x and x. In the second position of each binomial, we’re looking for the factors of -56 that add to -1, which are -8 and positive 7. So we have 0 = x - 8 times x + 7, which means that either 0 = x – 8 or 0 = x + 7. Finally, in the first equation, we add 8 to both sides, to get 8 = x. And in the second equation, we subtract 7 from both sides, to get -7 = x. So 8 = x or -7 = x. It’s important to understand that both of these answers work. Plugging an 8 back into the original problem, we have 8 is 56 less than 8 squared, or 8 = 8 squared – 56, which simplifies to 8 = 64 – 56, or 8 = 8, which is a true statement. And plugging a -7 back into the original problem, we have -7 is 56 less than -7 squared, or -7 = -7 squared – 56, which simplifies to -7 = 49 – 56, or -7 = -7, which is also a true statement.
Area | Maths Concept For Kids | Maths Grade 5 | Periwinkle
By Lumos Learning
Area | Maths Concept For Kids | Maths Grade 5 | Periwinkle
Customary Unit Conversions | MathHelp.com
By MathHelp.com
This lesson covers complex numbers. Students learn that a complex number is the sum or difference of a real number and an imaginary number and can be written in a + bi form. For example, 1 + 2i and -- 5 - i root 7 are complex numbers. Students then learn to add, subtract, multiply, and divide complex numbers that do not contain radicals, such as (5 + 3i) / (6 - 2i). To divide (5 + 3i) / (6 - 2i), the first step is to multiply both the numerator and denominator of the fraction by the conjugate of the denominator, which is (6 + 2i), then FOIL in both the numerator and denominator, and combine like terms.
Understand Quadrilaterals - Trapezoid, Rhombus, Parallelogram, Rectangle, Square, Kite - [13]
By Math and Science
Quality Math And Science Videos that feature step-by-step example problems!
08 - Solving Exponential Equations - Part 1 - Solve for the Exponent
By Math and Science
Quality Math And Science Videos that feature step-by-step example problems!
Difference of Two Cubes | MathHelp.com
By MathHelp.com
To solve the given system of inequalities, we start by graphing the associated equation for each inequality. In other words, we graph y equals -1/5 x +1 and y equals 3x + 2. So, for the first inequality, we start with our y-intercept of positive 1, up 1 unit on the y-axis. From there, we take our slope of -1/5, so we go down 1 and to the right 5, and plot a second point. Now, notice that our inequality uses a “less than” sign. This means that we draw a dotted line connecting the points, rather than a solid line. It’s important to understand that if we have a greater than sign or a less than sign, we use a dotted line, and if we have a greater than or equal to sign or a less than or equal to sign, we use a solid line. Pay close attention to this idea when drawing your lines. Students often carelessly use a solid line when they should use a dotted one, and vice-versa. Next, let’s take a look at our second inequality, which has a y-intercept of positive 2, up 2 units on the y-axis. From there, we take our slope of 3, or 3 over 1, so we go up 3 and to the right 1, and plot a second point. And notice that this inequality uses a “greater than or equal to” sign, so we connect the points with a solid line, rather than a dotted line. Next, we need to determine which side of each of these lines to shade on the graph. To determine which side of our first line to shade, we use a test point on either side of the first line. The easiest test point to use is (0, 0), so we plug a zero into our first inequality for both x and y, and we have 0 is less than -1/5 times 0 + 1, which simplifies to 0 is less than 0 + 1, or 0 is less than 1. Notice that 0 is less than 1 is a true statement. This means that our test point, (0, 0), is a solution to the first inequality, so we shade in the direction of (0, 0) along our first boundary line. Next, we determine which side of our second line to shade by using a test point on either side of the second line, such as (0, 0). Plugging a zero into our second inequality for both x and y, we have 0 equal to 3 times 0 + 2, which simplifies to 0 equal to 0 + 2, or 0 equal to 2. Notice that 0 equal to 2 is a false statement. This means that our test point, (0, 0), is a not solution to the inequality, so we shade away from (0, 0) along our second boundary line. Finally, it’s important to understand that the solution to this system of inequalities is represented by the part of the graph where the two shaded regions overlap, which in this case is in the lower left. In other words, any point that lies in this part of the graph is a solution to the given system of inequalities. Note that the points along the dotted boundary line of this region are not solutions to the system, but the points along the solid boundary line of this region are solutions to the system.
