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how to solve the expanded notation 4 over 5

how to solve the expanded notation 4 over 5

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What is Expanded Notation? - Definition & Examples - Video ...


... What is Expanded Notation? ... Let's look at a number written in expanded notation: (5 x 10,000 ... We have over 95 college courses that prepare you to ...
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Quiz & Worksheet - Expanded Notation Method for Division ...


Print Expanded Notation Method for Division Worksheet 1. ... Solving using expanded notation ... Explore our library of over 55,000 lessons. Search.
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Basic Math: Expanded Notation - YouTube


Oct 17, 2011 · Basic Math: Expanded Notation ... Expanded Notation to Solve a 3-Digit by 3-Digit Addition ... ES 4 Math Division with Expanded Notation ...

Expanded Notation - Softschools.com


Expanded form or expanded notation is a helpful way to rewrite numbers in order to ... (4 x ) The expanded notation helps us to see that the 3 is in the tenths place ...
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Expanded Notation - Let me solve your Math Problems


Let me solve your Math Problems. Search this site. Percentages. ... In the above equation written in expanded notation, you will see that I didn't write down the ...
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expanded notation - Math N Stuff


Write 4 x 1000 + 2 x 100 + 6 x 10 + 3 in standard notation. 2. Write 5 x 10,000 + 2 x 100 + 7 x 10 in standard notation. 3. Write 5,238 in expanded notation. 4.
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Definition of Expanded Notation - Math Is Fun


Expanded Notation. Writing a number to show the value of each digit. ... For example: 4,265 = 4 x 1,000 + 2 x 100 + 6 x 10 + 5 x 1. See: Standard Notation. Search
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Grade 4-Expanded Notation Method for Multiplication - YouTube


Jul 21, 2013 · Using the expanded notation method to solve one digit by two digit and two digit by two digit multiplication.

Suggested Questions And Answer :


how to solve the expanded notation 4 over 5

4 over 5 is 0.8 that in expanded notation is just 0.8
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what is 654.321 in expanded notation using exponents

654.321  expanded notation is the same as scientific notation. 6.54321 * 10^(-2)  the exponent is negative due to the fact that we moved the decimal to the left and made the number smaller.
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how do you solve this problem using set builder notation x=3/8>-3

how do you solve this problem using set builder notation x=3/8>-3 The inequality is: x + 3/8 > -3 x > -3 – 3/8 = -27/8 x > -27/8 In set builder notation this is, { x | x < -27/8}
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solve x+2/9>-5 using a set builder notation?

x+2/9>5 solve this problem using a set builder notation The inequality is, x + 2/9 > 5 x > 5 - 2/9 = 43/9 x > 43/9 In set builder notation this is: { x | x > 43/9}
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how do you use brackets in simultaneous equations

One reason you might use brackets in simultaneous equations is for expressing the solution as an ordered set (usually pairs). If, for example, you had a 2-variable system of equations, x and y, you might express the result as the ordered pair (x,y) where x is replaced by the value found for x and y the value found for y. Sometimes a system of equations has more than one solution, so the different solutions would be represented as (x1,y1) and (x2,y2). This can happen when the system involves one or more quadratic equations. The brackets ensure that the values for x and y are not mixed up. Another reason for brackets is when using substitution to solve a system. Suppose there are two equations: ax+by=c and dx+ey=f. From the first equations we can write y=(c-ax)/b and substitute for y in the second equation: dx+e(c-ax)/b=f. That would be the first step in solving. The next step would be to expand the brackets and solve for x in terms of the constants a, b, c, d, e and f. Having found x, you find y by substituting the value of x in y=(c-ax)/b. Sometimes a question involving simultaneous equations comes in a form that uses brackets, so the first step is to expand the brackets, combine like terms, and then continue to solve the system.
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How to solve a set with '...'s on both sides such as, {...,-1,0,1,...} and covert into set notation?

How to solve a set with '...'s on both sides such as, {...,-1,0,1,...} and covert into set notation? The set shown is an infinite set, the ellipses mean "and so on" down to - infinity and up to plus infinity. What you have there is the set of all integers, positive and negative. In set builder notatoin, ths would be {x | x e Z}, where Z is the set of all integers.
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4>8-3x/5>-2 I am having trouble solving this using interval notation.

Question: Solve 4>8-3x/5>-2. Split the above inequality into two inequalities. 4 > 8 - 3x/5 and 8 - 3x/5 > -2 Working both inequalities together, -4 > -3x/5    and   10 > 3x/5 -20 > -3x      and    50 > 3x 20/3 < x      and     x < 50/3 fitting the two inequalities together, 20/3 < x < 50/3 In set notation: x є  (20/3,50/3}
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sqrt(3a+10) = sqrt(2a-1) + 2

sqrt(3a+10)=sqrt(2a-1)+2 To remove the radical on the left-hand side of the equation, square both sides of the equation. (~(3a+10))^(2)=(~(2a-1)+2)^(2) Simplify the left-hand side of the equation. 3a+10=(~(2a-1)+2)^(2) Squaring an expression is the same as multiplying the expression by itself 2 times. 3a+10=(~(2a-1)+2)(~(2a-1)+2) Multiply each term in the first group by each term in the second group using the FOIL method. FOIL stands for First Outer Inner Last, and is a method of multiplying two binomials.  First, multiply the first two terms in each binomial group.  Next, multiply the outer terms in each group, followed by the inner terms.  Finally, multiply the last two terms in each group. 3a+10=(~(2a-1)*~(2a-1)+~(2a-1)*2+2*~(2a-1)+2*2) Simplify the FOIL expression by multiplying and combining all like terms. 3a+10=(~(2a-1)^(2)+4~(2a-1)+4) Remove the parentheses around the expression ~(2a-1)^(2)+4~(2a-1)+4. 3a+10=~(2a-1)^(2)+4~(2a-1)+4 Raising a square root to the square power results in the expression inside the root. 3a+10=(2a-1)+4~(2a-1)+4 Add 4 to -1 to get 3. 3a+10=2a+3+4~(2a-1) Since a is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation. 2a+3+4~(2a-1)=3a+10 Move all terms not containing ~(2a-1) to the right-hand side of the equation. 4~(2a-1)=-2a-3+3a+10 Simplify the right-hand side of the equation. 4~(2a-1)=a+7 Divide each term in the equation by 4. (4~(2a-1))/(4)=(a)/(4)+(7)/(4) Simplify the left-hand side of the equation by canceling the common terms. ~(2a-1)=(a)/(4)+(7)/(4) To remove the radical on the left-hand side of the equation, square both sides of the equation. (~(2a-1))^(2)=((a)/(4)+(7)/(4))^(2) Simplify the left-hand side of the equation. 2a-1=((a)/(4)+(7)/(4))^(2) Combine the numerators of all expressions that have common denominators. 2a-1=((a+7)/(4))^(2) Expand the exponent of 2 to the inside factor (a+7). 2a-1=((a+7)^(2))/((4)^(2)) Expand the exponent 2 to 4. 2a-1=((a+7)^(2))/(4^(2)) Simplify the exponents of 4^(2). 2a-1=((a+7)^(2))/(16) Multiply each term in the equation by 16. 2a*16-1*16=((a+7)^(2))/(16)*16 Simplify the left-hand side of the equation by multiplying out all the terms. 32a-16=((a+7)^(2))/(16)*16 Simplify the right-hand side of the equation by simplifying each term. 32a-16=(a+7)^(2) Since (a+7)^(2) contains the variable to solve for, move it to the left-hand side of the equation by subtracting (a+7)^(2) from both sides. 32a-16-(a+7)^(2)=0 Squaring an expression is the same as multiplying the expression by itself 2 times. 32a-16-((a+7)(a+7))=0 Multiply -1 by each term inside the parentheses. 32a-16-a^(2)-14a-49=0 Since 32a and -14a are like terms, add -14a to 32a to get 18a. 18a-16-a^(2)-49=0 Subtract 49 from -16 to get -65. 18a-65-a^(2)=0 Move all terms not containing a to the right-hand side of the equation. -a^(2)+18a-65=0 Multiply each term in the equation by -1. a^(2)-18a+65=0 For a polynomial of the form x^(2)+bx+c, find two factors of c (65) that add up to b (-18).  In this problem -5*-13=65 and -5-13=-18, so insert -5 as the right hand term of one factor and -13 as the right-hand term of the other factor. (a-5)(a-13)=0 Set each of the factors of the left-hand side of the equation equal to 0. a-5=0_a-13=0 Since -5 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 5 to both sides. a=5_a-13=0 Set each of the factors of the left-hand side of the equation equal to 0. a=5_a-13=0 Since -13 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 13 to both sides. a=5_a=13 The complete solution is the set of the individual solutions. a=5,13
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22 base 3 into expanded notation is?

????????????? "expanded notashun" ?????????? 22 in base 3=2*3+2 =8 in base 10
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What is (x-6)(x-6) in expanded notation?

(x-6)^2=x^2 -12x +36 .................
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