Explanation Step 1 1 of 3 x 2 b x − 24 x^2bx24 x 2 b x − 24 The given expression is completely factorizable if we can find two integers such that their sum is b b b and their product is 24 Following are the total number of ways in which we can write 24 as a product of two numbers 23 Solving Linear Equations Part 1 Identify linear equations with one variable and verify their solutions Use the properties of equality to solve basic linear equations Use multiple steps to solve linear equations by isolating the variable Solve linear equations where the coefficients are fractions or decimals Misc 5 Solve the inequality −12 < 4 −3𝑥/(−5) ≤ 2 −12 < 4 −3𝑥/(−5) ≤ 2 Subtracting 4 all sides (Eliminating 4) – 12 – 4 < 4 − 3𝑥/(−5
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3x+5-2(x+2)=8-3(1-x)
3x+5-2(x+2)=8-3(1-x)-SOLUTION by C2→ C2−C1 C3→ C3−4C1 =−(−x−2x)1(−2x−43x)x(2−3) =2x−4−x= −2 Trick Put x = 1 Then ∣∣ ∣ ∣2 3 5 4 6 9 8 11 15∣∣ ∣ ∣= −2Quadratic polynomial can be factored using the transformation a x 2 b x c = a (x − x 1 ) (x − x 2 ), where x 1 and x 2 are the solutions of the quadratic equation a x 2 b x c = 0 3x^{2}5x1=0



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D) −5sin(5x)5ex1 − 14 3 3 √ x−1; Solve the compound inequality Graph the solution and write the solution in interval notation 1 5x − 5 ≥ − 3 and − 4(x − 1) ≥ − 2 Answer Sometimes we have a compound inequality that can be written more concisely For example, a < x and x < b can be written simply as a < x < b and then we call it a double inequality5 a 5 6 a 1 2 − = 6 x 3 x x 1 x 3x 1 2 − = − − D Absolute value 1 5−2z −1=8 2 x5 −7 =−2 3 5x −1 =−2 4 4 1 4 3 x 2 1 − = 5 y −1 =7 y E Exponential 1 10x =1000 2 103x5 =100 3 8 1 2x1 = 4 () 3 1 3x2 9x = 5 () 8 1 2 2 42x = F Logarithmic 1 log2 ( )x 5 =log2 1−5x 2 2log3 ()x 1 =log3 4x 3 log2 ()x
Step 2 Use the appropriate properties of equality to combine oppositeside like terms with the variable term on one side of the equation and the constant term on the other Step 3 Divide or multiply as needed to isolate the variable Step 4 Check to see if the answer solves the original equationThe sum of these coefficients is 0 and thus the elimination method can be applied directly Step 3 3 of 5 Next, we subtract the two equations This will eliminate \textbf {eliminate} eliminate one of the variables We solve the equation to the other (uneliminated) variable ( 3 x − 2 y) − ( 2 x − 2 y) = 8 − 5 (3x2y) (2x2y)=85Click here👆to get an answer to your question ️ Find the value of the polynomial 3x^3 4x^2 7x 5, when x = 3 and also when x = 3 Join / Login Question Find the value of the polynomial 3 x 3 = 8 1 − 3 6 2 1 − 5 = 6 1
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Answer to − 5(x 2) 8x = − 2 3x − 8 Find solutions for your homework or get textbooks SearchA) (−5, −2) B) (3, −8) C) (4, −6) D) (9, −6) 534 views Share FollowX∞ n=1 5−2 √ n n3 We can temporarily break this apart to see if the pieces converge X∞ n=1 5−2 √ n n3 = X∞ n=1 5 n3 −2 X∞ n=1 √ n n3 Both of these are p−series, the first with p = 3, the second with p = 5 2, therefore they converge separately, and so the sum also converges 5 X∞ n=1 (−6)n−151−n First, let's



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A 3x 3 −10 B −3x 2 x−7 C 3x 3 −6x 2 5x−10 D 3x 2 −x7 Given f(x)=2x 2 −5x−3 and g(x)=2x 2 x What is (fg)(x) ?A x 8 B 3x 8 C x − 2 D x − 8 2 See answers Advertisement Advertisement Transcript Ex 93, 4 (a) (a) Simplify 3x (4x – 5) 3 and find its values for (i) x = 3 (ii) x = 1/2 3𝑥 (4𝑥−5)3 = (3𝑥×4𝑥)−(3𝑥×5)3 = 12𝑥^2−15𝑥3 (i) For 𝒙=𝟑 Putting 𝑥=3 in expression 12𝑥^2−15𝑥3 = 12(3)^2−15(3)3 = (12×9)−(15×3) 3 = 108−45 3 = 108−42 = 66 (ii) For 𝒙=𝟏/𝟐 Putting 𝑥=1/2 in expression 12𝑥^2−15𝑥3



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2(42x)≥5x5 x≤−2 x≥−2 x≤3 x≥3 2 See answers Advertisement Advertisement carlenehagen carlenehagen 8 4x >_ 5x 5,,, 3 >_ x (>_ just means greater than our equal to) Advertisement Advertisement krislynnkat krislynnkat X is less then of equal to 3 Answer x = 0 For the functions f ( x) in exercises 6 10, determine whether there is an asymptote at x = a Justify your answer without graphing on a calculator 6) f ( x) = x 1 x 2 5 x 4, a = − 1 7) f ( x) = x x − 2, a = 2 Answer Yes, there is a vertical asymptote at x = 2 8) f ( x) = ( x 2) 3 / 2, a = − 2Combine 4 x 2 and − 5 x 2 to get − x 2 Add 9 and to get 29 Add 9 and 2 0 to get 2 9 Multiply 2x3 and 2x3 to get \left (2x3\right)^ {2} Multiply 2 x − 3 and 2 x − 3 to get ( 2 x − 3) 2 Use binomial theorem \left (ab\right)^ {2}=a^ {2}2abb^ {2} to expand \left (2x3\right)^ {2}



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3cos(5x)2ln(4x)8 √ x 5 f) cos(4x)3ln(3x)−7 5 √ x 3 g) 4sin(3x)3ln(x)− √6 x −1 2 h) 4sin(5x)e3x − 4 √ 4 x 5 2 Answers a) 25cos(5x)16e4x1 − 2 x 3/2;Explanation Step 1 1 of 2 We are given a system of two linear equations We write each of the equations in a slopeintercept form, y = k x b y=kxb y = k x b 3 x − 2 y = 8 3 x − 2 y − 3 x = 8 − 3 x − 2 y = − 3 x 8 y = 3 2 x − 4 \begin {gather*} \color {#c} 3x2y=8\\ \color {#c} 3x2y3x=x \\ \color {#See the answer (a) f (x) = 3x 5 √ x − 12 √4 x 3 x 3 x (b) y = (3x ^6 − 2x ^3 8x ^2 5) (x ^5 − x ^2 2x − 8) (c) f (t) = t^7 3t − 5/ t ^2 − 5t (d) P (x) = 7/6 (2x ^5 − 3x − 4)^12 Find the first and second derivatives of each



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Ex 23, 3 Solve the following equations and check your results 5x 9 = 5 3x5x 9 = 5 3x 5x = 3x 5 − 9 5x = 3x − 4 5x − 3x = −4 2x = −4 x = (−4)/2 x = −2 Check LHS 5x 9 = 5 (−2) 9 = −10 9 = −1 RHS 5 3x = 5 3 (−2) = 5 − 6 = −1 ∴ LHS = RHS Hence Verif2 x 3 5 = x − 9 2 \frac {2x} {3}5= x\frac {9} {2} 32x 5 = x− 29 See answer › Systems of equations 1 Solve the system 5 x − 3 y = 6 4 x − 5 y = 12Solve for x and y7y3−2x2=144y−23x−3=2 The given sy Solve for x and y 7(y3)−2(x2)= 14,4(y−2)3(x−3) =2 Please scroll down to see the correct answer and solution guide



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E) 5−15sin(5x) 2 x √4 3cos(5x)2ln Example 1 Evaluate the expression x 2 − 2 x y y 2 at x = − 3 and y = 2 Solution Following "Tips for Evaluating Algebraic Expressions," first replace all occurrences of variables in the expression x2 − 2 xy y2 with open parentheses x 2 − 2 x y y 2 = ( ) 2 − 2 ( ) ( ) ( ) 2 Secondly, replace each variable with its givenSolutions for Chapter 85 Problem 8PS If the functions f, g, and h are defined by f (x) = 3x − 5, g(x) = x − 2, and h(x) = 3x2, write a formula for each of the following functions Examples 1–2f − g Get solutions Get solutions Get solutions done loading Looking for the textbook?



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X 1 − 2x 2 3x 3 x 4 = −3 2x 1 − x 2 3x 3 − x 4 = 0 Solution The augmented matrix of the given system is 1 −2 3 1 −3 2 −1 3 −1 0 A corresponding rowechelon matrix is obtained by adding negative two times the first row to the second row 1 −2 3 1 −3 0 3 −3 −3 6 Thus x 3 = s and x 4 = t are free variables Solving What's the simplified form of 2x 3 − x 5?Steps Using Factoring By Grouping 3 { x }^ { 2 } 10x8 = 0 3 x 2 1 0 x − 8 = 0 To solve the equation, factor the left hand side by grouping First, left hand side needs to be rewritten as 3x^ {2}axbx8 To find a and b, set up a system to be solved To solve the equation, factor the left hand side by grouping



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Use the Shell Method to compute the volume V of the solid obtained by rotating the region enclosed by the graphs of the functions y = x^2, y = 8 − x^2,and x = 1/2 about the yaxis Here is how I set up the integral 2 pi Statistics The monthly incomes from a random sample of workers in a factory are shown below Ex 23, 6 Solve the following equations and check your results 8x 4 = 3 (x – 1) 78x 4 = 3(x − 1) 7 8x 4 = 3x − 3 7 8x 4 = 3x 4 8x = 3x 4 − 4 8x = 3x 0 8x − 3x = 0 5x = 0 x = 0 Check LHS 8x 4 = 8 × 0 4 = 4 RHS 3 (x − 1) 7 = 3 (0 − 1) 7 = −34 x 2 3 x − 5 x 0 The powers of the variable x in the above equation are 2 , 1 and 0 which are whole numbers Therefore, 4 x 2 3 x − 5 is a polynomial



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Let p(x) = x 3 − 3x 2 − 9x − 5 All the factors of 5 have to be considered These are ±1, ± 5 By trial method, p(−1) = (−1) 3 − 3(−1) 2 − 9(−1) − 5 = − 1 − 3 9 − 5 = 0 Therefore, x 1 is a factor of this polynomial Let us find the quotient on dividing x 3 3x 2 − 9x − 5 by x 1 By long division,See answer › Exponential and logarithmic functions Solve for x 3 e 3 x ⋅ e − 2 x 5 = 2 3e^ {3x} \cdot e^ {2x5}=2 3e3x ⋅e−2x5 = 2 See answer › Systems of equations 2 Solve the system 2 9 ⋅ x − 5 y = 1 9 4 5 ⋅ x 3 y = 23x−10x−4 − 2 > 0 Then multiply 2 by (x−4)/(x−4) 3x−10x−4 − 2 x−4x−4 > 0 Now we have a common denominator, let's bring it all together 3x−10 − 2(x−4)x−4 > 0 Simplify x−2x−4 > 0 Second, let us find "points of interest" At x=2 we have (0)/(x−4) > 0, which is a "=0" point, or root At x=4 we have (x−2



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Click here👆to get an answer to your question ️ Factorise 2x^3 3x^2 17x 30 Join / Login Question (x 2 x − 6) (2 x − 5) = (x 2 3 x − 2 xSolution Steps ( 2 x 5 ) ( 3 x 8 ) ( 2 x 5) ( 3 x − 8) Apply the distributive property by multiplying each term of 2x5 by each term of 3x8 Apply the distributive property by multiplying each term of 2 x 5 by each term of 3 x − 8 6x^ {2}16x15x40 6 x 2 − 1 6 x 1 5 x − 4 0X 3 (4− x 2)3/ dx = − 1 15 4− x2 5/2 2 0 = 32 15 (2) (Problem 54, Problem 14) Evaluate the volume integral (triple integral) of f(x,y,z) = x2 over S, where S is the solid bounded by the paraboloids z = x2 y2 and z = 8−x2 − y2 Solution =4 z=8−x2−y2 z=x2y2 R S x2 y2 Figure 1 Region S bounded above by paraboloid z = 8−x2−



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2 5x 1 −x 2 = −3x 1 −3x 2 ⇒ 2x 1 2x 2 = −3x 1 and 5x 1 −x 2 = −3x 2 ⇒ 5x 1 = −2x 2 ⇒ x 1 = − 2 5 x 2 This means that, while there are infinitely many nonzero solutions (solution vectors) of the equation Ax = −3x, they all satisfy the condition that the first entry x 1 is −2/5 times the second entry x 2 Thus allCPT Review 4/17/01 5 Multiply 2x( 3x2 −5x −3) A) 6x3 −5x2 −6x B) 6x3 −5x −3 C) 6x3 −10x2 −3x D) 6x3 −10x2 −6x 21 Divide m m m m 7 14 2 − 28 8 7 A) 2m −28m8 7m B) 2m −4m7 1 C) 2 −4m 7 D) 2m2 −4m8 m 2233 Answers Absolute Value Inequalities 1) − 3, 3 2) − 8, 8 3) − 3, 3 4) − 7, 1 5) − 4, 8 6) − 4, 7) − 2, 4 8) − 7, 1 9) − 7 3, 11 3 10) − 7, 2



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