Algebra+2+Quarter+3+Review+Assignment

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Apr 25, 2024

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Algebra 2 Third Quarter Review Assignment Each problem below is similar to the questions on the 3 rd Quarter Practice Exam. You must show your work to receive credit! Question 1) Identify the inverse of the function. f ( x ) = 3 x + 2 4 1. Replace ( ) f ( x ) with y : =3 +24 y =43 x +2 . 2. Swap x and y : =3 +24 x =43 y +2 . 3. Solve for y . Starting from step 2: =3 +24 x =43 y +2 Multiply both sides by 4 to get rid of the denominator: 4 =3 +24 x =3 y +2 Subtract 2 from both sides: 4 −2=3 4 x −2=3 y Finally, divide both sides by 3 to solve for y : =4 −23 y =34 x −2 So, the inverse function of ( )=3 +24 f ( x )=43 x +2 is: −1( )=4 −23 f −1( x )=34 x −2 Question 2) Solve the equation. 27 x + 5 = 9 2 x 4 To solve the equation 27 +5=92 −427 x +5 =9 2 x −4 , notice that both 27 and 9 can be expressed as powers of 3 since 27=3327=3 3 and 9=329=3 2 . Rewriting the equation using this insight: 33( +5)=32(2 −4)3 3( x +5) =3 2(2 x −4) Simplifying each side: 33 +15=34 −83 3 x +15 =3 4 x −8 Since the bases are the same and the equation is set, the exponents must be equal: 3 +15=4 −83 x +15=4 x −8 Now, solve for x : 15+8=4 −3 15+8=4 x −3 x 23= 23= x Therefore, the solution to the equation 27 +5=92 −427 x +5 =9 2 x −4 is =23 x =23 .
Question 3) Solve the equation. lo g 3 ( 4 x 2 ) lo g 3 2 = 2 1. Combine the logarithmic expressions using the property that log �� −log �� =log ( �� )log bA −log b B =log b ( BA ): log3(4 −22)=2log3(24 x −2)=2 2. Simplify the fraction inside the logarithm: log3(2 −1)=2log3(2 x −1)=2 3. To get rid of the logarithm, convert the equation into its exponential form. This step is based on the definition of logarithms: if log �� = log bA = C , then �� = bC = A : 32=2 −132=2 x −1 4. Solve the resulting equation: 32=9=2 −132=9=2 x −1 Adding 1 to both sides gives: 9+1=2 9+1=2 x 10=2 10=2 x Divide both sides by 2 to solve for x : =102=5 x =210=5 Therefore, the solution to the equation log3(4 −2)−log32=2log3(4 x −2)−log32=2 is =5 x =5. Question 4) Solve the equation. x 21 x =− 4 To solve the equation −21 =−4 x x 21 =−4, it's convenient to start by getting rid of the fraction. First, let's rewrite the equation to see it clearly: −21 =−4 x x 21 =−4 Multiply every term by x to eliminate the fraction: 2−21=−4 x 2−21=−4 x Rearrange the equation to set it to 0, moving all terms to one side (ideal for solving quadratic equations): 2+4 −21=0 x 2+4 x −21=0 Now, we have a quadratic equation in the form of �� 2+ �� + =0 ax 2+ bx + c =0. To solve it, you can either factorize it (if possible), complete the square, or use the quadratic formula. The quadratic formula is: =− ± 2−4 �� 2 x =2 a b ± b 2−4 ac For our equation, =1 a =1, =4 b =4, and =−21 c =−21. Plugging these into the quadratic formula gives: =−4±42−4(1)(−21)2(1) x =2(1)−4±42−4(1)(−21) =−4±16+842 x =2−4±16+84 =−4±1002 x =2−4±100 =−4±102 x =2−4±10 This gives us two possible solutions for x : 1. =−4+102=62=3 x =2−4+10 =26 =3 2. =−4−102=−142=−7 x =2−4−10 =2−14 =−7 Therefore, the solutions to the equation −21 =−4 x x 21 =−4 are =3 x =3 and =−7 x =−7.
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