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Insert grouping symbols in the expression so thatthe balue of the expression is 14. 2x2+3^2-4+3x5

I really need help on this soon. Thanks!

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use of grouping symbols - FreeMathHelp.com


The assignment is to "insert grouping symbols in an expression to make it ... 2x2+3^2-4+3x5....would be (2x2+3)^2-(4+3)x5 = 14 So there are ... use of grouping ...
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I need to insert grouping symbols in the expression so that ...


... 58.Insert grouping symbols in the expression 2.3³ + 4 to ... so answer = 14. 2x2+3 to 2nd power-4+3x5. ... a value equal to half ... Math Insert grouping ...
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insert grouping symbols into the expression 4 plus 3 times 2 ...


I need to insert grouping symbols in the expression so that the value of the expression is 14 2x2+3 ... get 14 math insert grouping symbols into 2*3^3+4 to ...
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Grouping Symbols in Expressions - HelpingWithMath.com


Using Grouping Symbols in ... Parentheses are the most common grouping symbols. Grouping symbols in ... In the first expression, there are no parentheses, so solve ...
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Place grouping symbols in the expression so that it has a ...


Place grouping symbols in the expression so that it has a value of 14. 6 • 4 – 4 + 9 – 3? ... (11+9+3) = 14 ... (2x2+3)^2-(4+3)x5 = 14 So there are two more ...
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Solved: Insert grouping symbols in the proper place(s) so ...


Answer to Insert grouping symbols in the proper place(s) so that the given value of the expression is obtained.5 − 3 · 2 ... Insert grouping symbols in the proper ...
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evaluate the expression: 3 to the 3rd power-(15-8)+ 4 x 5 ...


evaluate the expression: 3 to ... I need to insert grouping symbols in the expression so that the value of the expression is 14 2x2+3(to the 2nd power)-4+3x5 I ...
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Parentheses in Number Sentences - Everyday Math


Students use parentheses in number sentences ... Students practice using grouping symbols. ... Story 2 goes with the first expression since 4 + 3 stands for the ...
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Suggested Questions And Answer :


Insert grouping symbols in the expression so thatthe balue of the expression is 14. 2x2+3^2-4+3x5

Insert grouping symbols in the expression so thatthe balue of the expression is 14. 2x2+3^2-4+3x5 Answer: = ( 2 x 2 + 3 ) ^ 2 - [ ( 4 + 3 ) x 5 ] = ( 4 + 3 ) ^ 2 - [ 7 x 5 ] = ( 7 ^ 2 ) - 35 = 49 - 35 = 14  
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insert grouping symbol in the expression 2*3^3+4=4374

2*3^7=4374 so yu need 2*3^(3+4)
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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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Where do I insert the grouping symbols to make this equation true, 9 divided by 8 minus 6 divided by 3 equals 6. or is it not possible?

me dont thank it be possabel No mater how yu mae put parens around parts, yu stil start with (9/3)...3 the 2 parts in the middel hav nuther divide..."divided bi 8" yu mite kobine (8-6), tu divide bi 2 insted av 8, but that make anser smaller If yu start with (9/8), get even smaller Noweer du yu hav add or multipli, so kant offset NE divide
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Rewrite the expression without using grouping symbols. 12(x 17)

Given 12(x 17) 204x
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solve equation 6x+5y=0, 3x-2y = 19

Exponents Exponents are supported on variables using the ^ (caret) symbol. For example, to express x2, enter x^2. Note: exponents must be positive integers, no negatives, decimals, or variables. Exponents may not currently be placed on numbers, brackets, or parentheses. Parentheses and Brackets Parentheses ( ) and brackets [ ] may be used to group terms as in a standard equation or expression. Multiplication, Addition, and Subtraction For addition and subtraction, use the standard + and - symbols respectively. For multiplication, use the * symbol. A * symbol is not necessiary when multiplying a number by a variable. For instance: 2 * x can also be entered as 2x. Similarly, 2 * (x + 5) can also be entered as 2(x + 5); 2x * (5) can be entered as 2x(5). The * is also optional when multiplying with parentheses, example: (x + 1)(x - 1). Order of Operations The calculator follows the standard order of operations taught by most algebra books - Parentheses, Exponents, Multiplication and Division, Addition and Subtraction. The only exception is that division is not currently supported; attempts to use the / symbol will result in an error. Division, Square Root, Radicals, Fractions The above features are not supported at this time. A future release will add this functionality.
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Where to insert grouping symbols to make this equation true 7+3*2+3=35

7 + 3 * 2 + 3 = 16 (7 + 3) * 2 + 3 = 23 7 + 3 * (2 + 3) = 22 (7 + 3) * (2 + 3) = 50 You might want to check the problem in your book and make sure the posted version is correct.
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Where do I insert the grouping symbols to make this equation true? 10 + 6 ÷ 2 - 3 = 5

(16/2)-3=5 ................
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how can you write two increased by six times a number in algebraic expression

  how can you write two increased by six times a number in algebraic expression 2 + 6n
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how to expand and simplify -4l-8(5m+3l)

big quesshun...wot is "i"...iz that stuf spozed tu be KOMPLEX NUMBERS???????
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