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Dividing Scientific Notation | MathHelp.com
By MathHelp.com
To multiply numbers that are in written in scientific notation, such as 1.4 x 10 to the -2nd times 5.3 times 10 to the 6th, we first multiply the decimals, in this case 1.4 times 5.3, to get 7.42. Next, we multiply the powers of 10, in this case 10 to the -2nd times 10 to the 6th. Notice that we’re multiplying two powers that have like bases, so we add the exponents and leave the base the same, to get 10 to the -2 + 6, or 10 to the 4th. So we have 7.42 times 10 to the 4th. Finally, we’re asked to write our answer in scientific notation. Notice, however, that 7.42 times 10 to the -4th is already written in scientific notation, because we have a decimal between 1 and 10 that is multiplied by a power of 10. So we have our answer.
GRE Math Test Prep | MathHelp.com
By MathHelp.com
This lesson covers the product rule. Students learn the product rule, which states that when multiplying two powers that have the same base, add the exponents. For example, x^4 times x^3 = x^7. To multiply 6s^3 times 3s^6, multiply the coefficients and add the exponents, to get 18s^9. If there is no exponent on the variable, it can be given an exponent of 1. For example, x can be thought of as x^1.
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.
Multiplying Integers | MathHelp.com
By MathHelp.com
This lesson covers comparing fractions. Students learn to compare fractions with the same denominator, which are called like fractions, by comparing the numerators. For example, to compare 7/9 and 4/9, note that 7 is greater than 4, so 7/9 is greater than 4/9. Students also learn to compare fractions with the different denominators, which are called unlike fractions, by first finding a common denominator, then comparing the numerators. For example, to compare 1/2 and 1/3, first find a common denominator, or the Least Common Multiple of 2 and 3, which is 6. To get 6 in the denominator of 1/2, multiply the numerator and denominator by 3, to get 3/6. To get 6 in the denominator of 1/3, multiply the numerator and denominator by 2, to get 2/6. Next, compare 3/6 and 2/6. Note that 3 is greater than 2, so 3/6 is greater than 2/6, which means that 1/2 is greater than 1/3.
6.RP.3 - Solve Word Problems Using Tape Diagrams
By MathwithMrAlmeida
In this video, Mr. Almeida shows you how to solve ratio word problems using tape diagrams. The models create equivalent ratios because you are multiplying both quantities in one ratio by the same positive number (the length of 1 unit). This meets Common Core Standard for Mathematics 6.RP.3.
Geometry Basics | MathHelp.com
By MathHelp.com
This lesson covers permutations. Students learn that a permutation is an arrangement of objects in which the order is important. For example, the permutation AB is different than the permutation BA. Students are then asked to solve word problems involving permutations. For example: Find the number of different ways 6 books can be arranged on a shelf. Note that the number of permutations can be found by multiplying the number of choices for the 1st position (6 books) times the number of choices for the second position (5 books), and so on. So the number of permutations is 6 x 5 x 4 x 3 x 2 x 1, or 720. In other words, there are 720 different ways 6 books can be arranged on a shelf.
Negative Exponents | MathHelp.com
By MathHelp.com
In this example, we’re given the functions f(x) = 3x – 2 (read as “f of x equals…”) and g(x) = root x, and we’re asked to find the composite functions f(g(9)) (read as “f of g of 9”) and g(f(9). To find f(g(9)), we first find g(9). Since g(x) = root x, we can find g(9) by substituting a 9 in for the x in the function, to get g(9) = root 9, and the square root of 9 is 3, so g(9) = 3. Now, since g(9) = 3, f(g(9)) is the same thing as f(3), so our next step is to find f(3). And remember that f(x) = 3x – 2, so to find f(3), we substitute a 3 in for the x in the function, and we have f(3) = 3 times 3 minus 2. Notice that I always use parentheses when substituting a value into a function, in this case 3. Finally, 3 times 3 minus 2 simplifies to 9 minus 2, or 7, so f(3) = 7. Therefore, f(g(9)) = 7. Next, to find g(f(9), we first find f(9). Since f(x) = 3x - 2, we find f(9) by substituting a 9 in for the x in the function, to get f(9) = 3 times 9 minus 2, which simplifies to 27 – 2, or 25, so f(9) = 25. Now, since f(9) = 25, g(f(9)) is the same thing as g(25), so our next step is to find g(25). And remember that g(x) = root x, so to find g(25), we substitute a 25 in for the x in the function, to get g(25) = root 25. Finally, the square root of 25 is 5, so g(25) = 5. Therefore, g(f(9)) = 5. It’s important to recognize that
Transitioning from counting to multiplying to find area | 3rd grade | Khan Academy
By Khan Academy
finds the area of a rectangle both by counting unit squares and multiplying side lengths
Synthetic Division | MathHelp.com
By MathHelp.com
In this example, it’s tempting to divide x squared + 5x – 6 by x + 1 by first factoring x squared + 5x – 6. The factors of -6 that add to positive 5 are +6 and -1, so we have x + 6 times x – 1 over x + 1. Notice, however, that nothing cancels. In this situation, we need a different method of dividing the polynomials, so we use long division. In other words, we rewrite x squared + 5x – 6 divided by x + 1 as x + 1 divided into x squared + 5x – 6. Now, our first step in the long division is to determine how many times x goes into x squared. Since x goes into x squared x times, we write an x above the x squared, just like we do with regular long division. Next, we multiply the x times the x + 1 in the divisor to get x squared + x, and we write the x squared + x underneath the x squared + 5x. Next, we subtract x squared + x from x squared + 5x. And watch out for this step: it’s an area where most of the common mistakes in these types of problems are made. Instead of subtracting, I would change the sign of each term in x squared + x, so we have negative x squared + negative x, then add the columns. So we have x squared + negative x squared, which cancels out, and positive 5x + negative x, which is positive 4x. Next, we bring down the -6, in regular long division. Now, we need to determine how many times x goes into 4x. Since x goes into 4x 4 times, we write a positive 4 in our answer. Next, we multiply positive 4 times x + 1 to get 4x + 4, and we write the 4x + 4 underneath the 4x – 6. Next, we subtract 4x + 4 from 4x – 6. In other words, we change the signs on 4x + 4 to -4x + -4, and we add. 4x + -4x cancels out, and -6 + -4 is -10. And since there are no other numbers to bring down, we have a remainder of -10. Finally, remember from the previous example that we add the remainder over the divisor to the quotient. In other words, we add -10 over x + 1 to x + 4, and we have x + 4 + -10 over x + 1. So x squared + 5x – 6 divided by x + 1 simplifies to x + 4 + -10 over x + 1.
Understand & Calculate Equivalent Fractions - [11]
By Math and Science
Quality Math And Science Videos that feature step-by-step example problems!
Multiplying decimals 3
By Khan Academy
Sometimes multiplying really small decimals (with all those zeros!) can be a little intimidating. Watch as we show you a handy trick to simplify these problems and solve them.
Multiplying decimals 3
By Khan Academy
Multiplying decimals? Try multiplying without the decimals first, them add them back in. We'll show you.
More involved multiplying decimals example
By Khan Academy
Sometimes multiplying really small decimals (with all those zeros!) can be a little intimidating. Watch as we show you a handy trick to simplify these problems and solve them.
Multiplying fractions by fractions word problems
By Khan Academy
You know how long it takes to bike to your friend's house, and you know how fast your can ride. So how far away does your friend live? The answer is one fraction multiplication problem away!