Guide :

2a-[5a^2-(4a-4a^2+14)]

add or subtract the polynomials, gather like terms, and write the simplified expression in standard form.

Research, Knowledge and Information :


What is the value of a? 14 – 2a = 4a – 16 - Brainly.com


What is the value of a? 14 – 2a = 4a – 16 2. Ask for details ; Follow; Report; by howdeshell 09/29/2015. Log in to add a comment ... 14 - 2x = 4a - 16 First, add ...
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14-2a=4a-16 - solution - Get Easy Solution


Simple and best practice solution for 14-2a=4a-16 equation. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, ...
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a^3-2a^2-3a^2-4a^3= - solution - Get Easy Solution


... -2a 2 + -3a 2 + a 3 + -4a 3 = 0 Combine like terms: -2a 2 + -3a 2 = -5a 2-5a 2 + a 3 + -4a 3 = 0 Combine like terms: ... =11 | | 2(4x-8)=8x+14 | | 3.2=5x+0.7 ...
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(a 3 - 2a + 5) - (4a 3 - 5a 2 + a - 2) - Brainly.com


(a^3 - 2a + 5) - (4a^3 - 5a^2 + a - 2) =a^3 - 2a + 5 ... I need help with 14 and 15 Mathematics; 5 points 27 minutes ago Which of the following are the factors of M^2 ...
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mod7grp2 - 2a 5 9 ab 2 2 a 5a 4(a 4(a 4a 3 a b 7ab 12b 2 2a 5 ...


... (a + 4a + 3) a b + 7ab + 12b 2 2a + 5 9 a + (a + 4)(a + 1) ... TERM Summer '14; ... Group 2 Challenge Problem 2a 5 9 ab 2 2 a 5a 4 ( a 4)( a 4a 3) a b 7ab 12b 2 ...
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5/(a+5) + 7/(2a-3) = 4a/ (2a^2+7a-15)? - Weknowtheanswer


Answer #2 | 14/10 2015 08:01 not sure, arctan(-11/2) = -2/(3/k) so k=2.1 or arctan ... (4a²+3a²)+(7a-2a)+3 =7a²+5a+3; ... 7a+4a=(7+4)a=11a. 2-2a+3a=2+a. 2+3=5.
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4A Top 25 (April 10) | Texas Highschool Baseball


14-5A; 15-5A; 16-5A; Region 3. 17-5A; 18-5A; 19-5A; 20-5A; ... 4A Top 25 (April 10) 1. Argyle (18-1-1) 2. ... 5A | 4A | 3A | 2A | Forums | Preseason | Playoffs |
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Football State Archives — University Interscholastic League (UIL)


Football State Archives. ... 21-14: Playoff Bracket: 2008-2009: 5A Division 2: Katy: Wylie: 17-3: ... 20-14: Playoff Bracket: 2007-2008: 4A Division 2:
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Suggested Questions And Answer :


Add 5a-3b+3a3=5a2+(-7a)-(4b)+5=


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5a2+32a-21

5x^2 +32x -21 yuze quadratik formula x=0.6 & -7 or (x-0.6)*(x+7) -b/2a=-3.2 b^2 -4ac=1444, root=38, root/2a=3.8
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find all solutions of the recurrence relation

n=2: a2=5a1+6a0+14; n=3: a3=5a2+6a1+21; etc. We do not have a0 or a1 to calculate a2 and further terms.   Assume the following: "note for  5a(n-1)  in this (n-1) is to the base   same for 6a(n-2)  in this n-2 is to the base  and for 7n  in this n is to the power " which needs to be interpreted. So does this mean: a[n]=5a[n-2]+6a[n-1]+7^n? The square brackets are meant to represent nth and other terms: a[0] means a0, the first term, for example. The second term is a1 or a[1]. If it does then: a2=5a0+6a1+7^2=5a0+6a1+49. If there are no terms before a0, a0=1, a1=6+7=13; a2=5+78+49=132; a3=65+792+343=1200; etc. The expression for a[n] can be split into 3 parts: 5a[n-2], 6a[n-1], 7^n. More...
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Rewriting the following form using the greatest con factor for (50a3 + 10a2) is what?

50x^3 +10x^2 bekum 10x^2(5x+1).....
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5a2-13a+13=7

5a^2-13a+13=7 5a^2-13+6=0 (5a-3)(a-2)=0 a=3/5;2
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How do you solve 5a(a-7)

5a(a-7) 5a2-35 5a2/5-35/5 a2=7
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