BIG IDEAS MATH Integrated Math 1: Student Edition 2016
16th Edition
ISBN: 9781680331127
Author: HOUGHTON MIFFLIN HARCOURT
Publisher: Houghton Mifflin Harcourt
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Chapter 1.5, Problem 7E
To determine
To calculate:
The value of
Expert Solution & Answer
Answer to Problem 7E
The value of
Explanation of Solution
Given information:
Calculation:
The given literal equation is:
Subtract
Multiply by
Chapter 1 Solutions
BIG IDEAS MATH Integrated Math 1: Student Edition 2016
Ch. 1.1 - Prob. 1ECh. 1.1 - Prob. 2ECh. 1.1 - Prob. 3ECh. 1.1 - Prob. 4ECh. 1.1 - Prob. 5ECh. 1.1 - Prob. 6ECh. 1.1 - Prob. 7ECh. 1.1 - Prob. 8ECh. 1.1 - Prob. 9ECh. 1.1 - Prob. 10E
Ch. 1.1 - Prob. 11ECh. 1.1 - Prob. 12ECh. 1.1 - Prob. 13ECh. 1.1 - Prob. 14ECh. 1.1 - Prob. 15ECh. 1.1 - Prob. 16ECh. 1.1 - Prob. 17ECh. 1.1 - Prob. 18ECh. 1.1 - Prob. 19ECh. 1.1 - Prob. 20ECh. 1.1 - Prob. 21ECh. 1.1 - Prob. 22ECh. 1.1 - Prob. 23ECh. 1.1 - Prob. 24ECh. 1.1 - Prob. 25ECh. 1.1 - Prob. 26ECh. 1.1 - Prob. 27ECh. 1.1 - Prob. 28ECh. 1.1 - Prob. 29ECh. 1.1 - Prob. 30ECh. 1.1 - Prob. 31ECh. 1.1 - Prob. 32ECh. 1.1 - Prob. 33ECh. 1.1 - Prob. 34ECh. 1.1 - Prob. 35ECh. 1.1 - Prob. 36ECh. 1.1 - Prob. 37ECh. 1.1 - Prob. 38ECh. 1.1 - Prob. 39ECh. 1.1 - Prob. 40ECh. 1.1 - Prob. 41ECh. 1.1 - Prob. 42ECh. 1.1 - Prob. 43ECh. 1.1 - Prob. 44ECh. 1.1 - Prob. 45ECh. 1.1 - Prob. 46ECh. 1.1 - Prob. 47ECh. 1.1 - Prob. 48ECh. 1.1 - Prob. 49ECh. 1.1 - Prob. 50ECh. 1.1 - Prob. 51ECh. 1.1 - Prob. 52ECh. 1.1 - Prob. 53ECh. 1.1 - Prob. 54ECh. 1.1 - Prob. 55ECh. 1.1 - Prob. 56ECh. 1.1 - Prob. 57ECh. 1.1 - Prob. 58ECh. 1.1 - Prob. 59ECh. 1.1 - Prob. 60ECh. 1.1 - Prob. 61ECh. 1.1 - Prob. 62ECh. 1.1 - Prob. 63ECh. 1.1 - Prob. 64ECh. 1.1 - Prob. 65ECh. 1.2 - Prob. 1ECh. 1.2 - Prob. 2ECh. 1.2 - Prob. 3ECh. 1.2 - Prob. 4ECh. 1.2 - Prob. 5ECh. 1.2 - Prob. 6ECh. 1.2 - Prob. 7ECh. 1.2 - Prob. 8ECh. 1.2 - Prob. 9ECh. 1.2 - Prob. 10ECh. 1.2 - Prob. 11ECh. 1.2 - Prob. 12ECh. 1.2 - Prob. 13ECh. 1.2 - Prob. 14ECh. 1.2 - Prob. 15ECh. 1.2 - Prob. 16ECh. 1.2 - Prob. 17ECh. 1.2 - Prob. 18ECh. 1.2 - Prob. 19ECh. 1.2 - Prob. 20ECh. 1.2 - Prob. 21ECh. 1.2 - Prob. 22ECh. 1.2 - Prob. 23ECh. 1.2 - Prob. 24ECh. 1.2 - Prob. 25ECh. 1.2 - Prob. 26ECh. 1.2 - Prob. 27ECh. 1.2 - Prob. 28ECh. 1.2 - Prob. 29ECh. 1.2 - Prob. 30ECh. 1.2 - Prob. 31ECh. 1.2 - Prob. 32ECh. 1.2 - Prob. 33ECh. 1.2 - Prob. 34ECh. 1.2 - Prob. 35ECh. 1.2 - Prob. 36ECh. 1.2 - Prob. 37ECh. 1.2 - Prob. 38ECh. 1.2 - Prob. 39ECh. 1.2 - Prob. 40ECh. 1.2 - Prob. 41ECh. 1.2 - Prob. 42ECh. 1.2 - Prob. 43ECh. 1.2 - Prob. 44ECh. 1.2 - Prob. 45ECh. 1.2 - Prob. 46ECh. 1.2 - Prob. 47ECh. 1.2 - Prob. 48ECh. 1.2 - Prob. 49ECh. 1.2 - Prob. 50ECh. 1.2 - Prob. 51ECh. 1.2 - Prob. 52ECh. 1.2 - Prob. 53ECh. 1.2 - Prob. 54ECh. 1.2 - Prob. 55ECh. 1.2 - Prob. 56ECh. 1.2 - Prob. 57ECh. 1.2 - Prob. 58ECh. 1.2 - Prob. 59ECh. 1.2 - Prob. 60ECh. 1.2 - Prob. 61ECh. 1.2 - Prob. 62ECh. 1.2 - Prob. 63ECh. 1.2 - Prob. 64ECh. 1.2 - Prob. 65ECh. 1.3 - Prob. 1ECh. 1.3 - Prob. 2ECh. 1.3 - Prob. 3ECh. 1.3 - Prob. 4ECh. 1.3 - Prob. 5ECh. 1.3 - Prob. 6ECh. 1.3 - Prob. 7ECh. 1.3 - Prob. 8ECh. 1.3 - Prob. 9ECh. 1.3 - Prob. 10ECh. 1.3 - Prob. 11ECh. 1.3 - Prob. 12ECh. 1.3 - Prob. 13ECh. 1.3 - Prob. 14ECh. 1.3 - Prob. 15ECh. 1.3 - Prob. 16ECh. 1.3 - Prob. 17ECh. 1.3 - Prob. 18ECh. 1.3 - Prob. 19ECh. 1.3 - Prob. 20ECh. 1.3 - Prob. 21ECh. 1.3 - Prob. 22ECh. 1.3 - Prob. 23ECh. 1.3 - Prob. 24ECh. 1.3 - Prob. 25ECh. 1.3 - Prob. 26ECh. 1.3 - Prob. 27ECh. 1.3 - Prob. 28ECh. 1.3 - Prob. 29ECh. 1.3 - Prob. 30ECh. 1.3 - Prob. 31ECh. 1.3 - Prob. 32ECh. 1.3 - Prob. 33ECh. 1.3 - Prob. 34ECh. 1.3 - Prob. 35ECh. 1.3 - Prob. 36ECh. 1.3 - Prob. 37ECh. 1.3 - Prob. 38ECh. 1.3 - Prob. 39ECh. 1.3 - Prob. 40ECh. 1.3 - Prob. 41ECh. 1.3 - Prob. 42ECh. 1.3 - Prob. 43ECh. 1.3 - Prob. 44ECh. 1.3 - Prob. 1QCh. 1.3 - Prob. 2QCh. 1.3 - Prob. 3QCh. 1.3 - Prob. 4QCh. 1.3 - Prob. 5QCh. 1.3 - Prob. 6QCh. 1.3 - Prob. 7QCh. 1.3 - Prob. 8QCh. 1.3 - Prob. 9QCh. 1.3 - Prob. 10QCh. 1.3 - Prob. 11QCh. 1.3 - Prob. 12QCh. 1.3 - Prob. 13QCh. 1.3 - Prob. 14QCh. 1.3 - Prob. 15QCh. 1.3 - Prob. 16QCh. 1.3 - Prob. 17QCh. 1.3 - Prob. 18QCh. 1.3 - Prob. 19QCh. 1.4 - Prob. 1ECh. 1.4 - Prob. 2ECh. 1.4 - Prob. 3ECh. 1.4 - Prob. 4ECh. 1.4 - Prob. 5ECh. 1.4 - Prob. 6ECh. 1.4 - Prob. 7ECh. 1.4 - Prob. 8ECh. 1.4 - Prob. 9ECh. 1.4 - Prob. 10ECh. 1.4 - Prob. 11ECh. 1.4 - Prob. 12ECh. 1.4 - Prob. 13ECh. 1.4 - Prob. 14ECh. 1.4 - Prob. 15ECh. 1.4 - Prob. 16ECh. 1.4 - Prob. 17ECh. 1.4 - Prob. 18ECh. 1.4 - Prob. 19ECh. 1.4 - Prob. 20ECh. 1.4 - Prob. 21ECh. 1.4 - Prob. 22ECh. 1.4 - Prob. 23ECh. 1.4 - Prob. 24ECh. 1.4 - Prob. 25ECh. 1.4 - Prob. 26ECh. 1.4 - Prob. 27ECh. 1.4 - Prob. 28ECh. 1.4 - Prob. 29ECh. 1.4 - Prob. 30ECh. 1.4 - Prob. 31ECh. 1.4 - Prob. 32ECh. 1.4 - Prob. 33ECh. 1.4 - Prob. 34ECh. 1.4 - Prob. 35ECh. 1.4 - Prob. 36ECh. 1.4 - Prob. 37ECh. 1.4 - Prob. 38ECh. 1.4 - Prob. 39ECh. 1.4 - Prob. 40ECh. 1.4 - Prob. 41ECh. 1.4 - Prob. 42ECh. 1.4 - Prob. 43ECh. 1.4 - Prob. 44ECh. 1.4 - Prob. 45ECh. 1.4 - Prob. 46ECh. 1.4 - Prob. 47ECh. 1.4 - Prob. 48ECh. 1.4 - Prob. 49ECh. 1.4 - Prob. 50ECh. 1.4 - Prob. 51ECh. 1.4 - Prob. 52ECh. 1.4 - Prob. 53ECh. 1.4 - Prob. 54ECh. 1.4 - Prob. 55ECh. 1.4 - Prob. 56ECh. 1.4 - Prob. 57ECh. 1.4 - Prob. 58ECh. 1.4 - Prob. 59ECh. 1.4 - Prob. 60ECh. 1.4 - Prob. 61ECh. 1.4 - Prob. 62ECh. 1.4 - Prob. 63ECh. 1.4 - Prob. 64ECh. 1.4 - Prob. 65ECh. 1.4 - Prob. 66ECh. 1.4 - Prob. 67ECh. 1.5 - Prob. 1ECh. 1.5 - Prob. 2ECh. 1.5 - Prob. 3ECh. 1.5 - Prob. 4ECh. 1.5 - Prob. 5ECh. 1.5 - Prob. 6ECh. 1.5 - Prob. 7ECh. 1.5 - Prob. 8ECh. 1.5 - Prob. 9ECh. 1.5 - Prob. 10ECh. 1.5 - Prob. 11ECh. 1.5 - Prob. 12ECh. 1.5 - Prob. 13ECh. 1.5 - Prob. 14ECh. 1.5 - Prob. 15ECh. 1.5 - Prob. 16ECh. 1.5 - Prob. 17ECh. 1.5 - Prob. 18ECh. 1.5 - Prob. 19ECh. 1.5 - Prob. 20ECh. 1.5 - Prob. 21ECh. 1.5 - Prob. 22ECh. 1.5 - Prob. 23ECh. 1.5 - Prob. 24ECh. 1.5 - Prob. 25ECh. 1.5 - Prob. 26ECh. 1.5 - Prob. 27ECh. 1.5 - Prob. 28ECh. 1.5 - Prob. 29ECh. 1.5 - Prob. 30ECh. 1.5 - Prob. 31ECh. 1.5 - Prob. 32ECh. 1.5 - Prob. 33ECh. 1.5 - Prob. 34ECh. 1.5 - Prob. 35ECh. 1.5 - Prob. 36ECh. 1.5 - Prob. 37ECh. 1.5 - Prob. 38ECh. 1.5 - Prob. 39ECh. 1.5 - Prob. 40ECh. 1.5 - Prob. 41ECh. 1.5 - Prob. 42ECh. 1.5 - Prob. 43ECh. 1.5 - Prob. 44ECh. 1.5 - Prob. 45ECh. 1.5 - Prob. 46ECh. 1.5 - Prob. 47ECh. 1.5 - Prob. 48ECh. 1.5 - Prob. 49ECh. 1.5 - Prob. 50ECh. 1.5 - Prob. 51ECh. 1.5 - Prob. 52ECh. 1.5 - Prob. 53ECh. 1.5 - Prob. 54ECh. 1 - Prob. 1CRCh. 1 - Prob. 2CRCh. 1 - Prob. 3CRCh. 1 - Prob. 4CRCh. 1 - Prob. 5CRCh. 1 - Prob. 6CRCh. 1 - Prob. 7CRCh. 1 - Prob. 8CRCh. 1 - Prob. 9CRCh. 1 - Prob. 10CRCh. 1 - Prob. 11CRCh. 1 - Prob. 12CRCh. 1 - Prob. 13CRCh. 1 - Prob. 14CRCh. 1 - Prob. 15CRCh. 1 - Prob. 16CRCh. 1 - Prob. 17CRCh. 1 - Prob. 18CRCh. 1 - Prob. 19CRCh. 1 - Prob. 20CRCh. 1 - Prob. 21CRCh. 1 - Prob. 22CRCh. 1 - Prob. 23CRCh. 1 - Prob. 1CTCh. 1 - Prob. 2CTCh. 1 - Prob. 3CTCh. 1 - Prob. 4CTCh. 1 - Prob. 5CTCh. 1 - Prob. 6CTCh. 1 - Prob. 7CTCh. 1 - Prob. 8CTCh. 1 - Prob. 9CTCh. 1 - Prob. 10CTCh. 1 - Prob. 11CTCh. 1 - Prob. 12CTCh. 1 - Prob. 13CTCh. 1 - Prob. 14CTCh. 1 - Prob. 15CTCh. 1 - Prob. 16CTCh. 1 - Prob. 1CACh. 1 - Prob. 2CACh. 1 - Prob. 3CACh. 1 - Prob. 4CACh. 1 - Prob. 5CACh. 1 - Prob. 6CACh. 1 - Prob. 7CACh. 1 - Prob. 8CACh. 1 - Prob. 9CA
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- For each real-valued nonprincipal character x mod k, let A(n) = x(d) and F(x) = Σ : dn * Prove that F(x) = L(1,x) log x + O(1). narrow_forwardBy considering appropriate series expansions, e². e²²/2. e²³/3. .... = = 1 + x + x² + · ... when |x| < 1. By expanding each individual exponential term on the left-hand side the coefficient of x- 19 has the form and multiplying out, 1/19!1/19+r/s, where 19 does not divide s. Deduce that 18! 1 (mod 19).arrow_forwardProof: LN⎯⎯⎯⎯⎯LN¯ divides quadrilateral KLMN into two triangles. The sum of the angle measures in each triangle is ˚, so the sum of the angle measures for both triangles is ˚. So, m∠K+m∠L+m∠M+m∠N=m∠K+m∠L+m∠M+m∠N=˚. Because ∠K≅∠M∠K≅∠M and ∠N≅∠L, m∠K=m∠M∠N≅∠L, m∠K=m∠M and m∠N=m∠Lm∠N=m∠L by the definition of congruence. By the Substitution Property of Equality, m∠K+m∠L+m∠K+m∠L=m∠K+m∠L+m∠K+m∠L=°,°, so (m∠K)+ m∠K+ (m∠L)= m∠L= ˚. Dividing each side by gives m∠K+m∠L=m∠K+m∠L= °.°. The consecutive angles are supplementary, so KN⎯⎯⎯⎯⎯⎯∥LM⎯⎯⎯⎯⎯⎯KN¯∥LM¯ by the Converse of the Consecutive Interior Angles Theorem. Likewise, (m∠K)+m∠K+ (m∠N)=m∠N= ˚, or m∠K+m∠N=m∠K+m∠N= ˚. So these consecutive angles are supplementary and KL⎯⎯⎯⎯⎯∥NM⎯⎯⎯⎯⎯⎯KL¯∥NM¯ by the Converse of the Consecutive Interior Angles Theorem. Opposite sides are parallel, so quadrilateral KLMN is a parallelogram.arrow_forwardBy considering appropriate series expansions, ex · ex²/2 . ¸²³/³ . . .. = = 1 + x + x² +…… when |x| < 1. By expanding each individual exponential term on the left-hand side and multiplying out, show that the coefficient of x 19 has the form 1/19!+1/19+r/s, where 19 does not divide s.arrow_forwardLet 1 1 r 1+ + + 2 3 + = 823 823s Without calculating the left-hand side, prove that r = s (mod 823³).arrow_forwardFor each real-valued nonprincipal character X mod 16, verify that L(1,x) 0.arrow_forward*Construct a table of values for all the nonprincipal Dirichlet characters mod 16. Verify from your table that Σ x(3)=0 and Χ mod 16 Σ χ(11) = 0. x mod 16arrow_forwardFor each real-valued nonprincipal character x mod 16, verify that A(225) > 1. (Recall that A(n) = Σx(d).) d\narrow_forward24. Prove the following multiplicative property of the gcd: a k b h (ah, bk) = (a, b)(h, k)| \(a, b)' (h, k) \(a, b)' (h, k) In particular this shows that (ah, bk) = (a, k)(b, h) whenever (a, b) = (h, k) = 1.arrow_forward20. Let d = (826, 1890). Use the Euclidean algorithm to compute d, then express d as a linear combination of 826 and 1890.arrow_forwardLet 1 1+ + + + 2 3 1 r 823 823s Without calculating the left-hand side, Find one solution of the polynomial congruence 3x²+2x+100 = 0 (mod 343). Ts (mod 8233).arrow_forwardBy considering appropriate series expansions, prove that ez · e²²/2 . e²³/3 . ... = 1 + x + x² + · ·. when <1.arrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
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