MATHEMATICS COLLEGE

Solve for x Write both solutions, seperated by a comma 4x^2+5x+1=0

Answers

Answer 1

Answer:

Answer:

-1/4 , -1

Step-by-step explanation:

I solved it using Factorization method and Quadratic Equation .

Factorization Method

$4x^2+5x+1=0\nWrite +5x- as- a -difference(write+5x-using- two- numbers -in-which-their-sum-is ; 5-and-their-product-is ; 4)\n4x^(2) +4x+1x+1=0\nFactorize-out-common-terms\n4x(x+1)+1(x+1)=0\nFactor-out-(x+1)\n(4x+1)(x+1)=0\n4x+1 =0 \nx+1=0\n4x=0-1\nx =0-1\n4x =-1\nx =-1\n4x=-(1)/(4) \n\nAnswer = -1/4 , -1$

Quadratic Equation

$4x^2+5x+1=0\na = 4\nb =5\nc = 1\n\nx =(-b\±√(b^2 -4ac) )/(2a) \n\nx = (-(5)\±√((5)^2-4(4)(1)) )/(2(4)) \n\nx = (-5\±√(25-16) )/(8) \n\nx = (-5\±√(9) )/(8) \n\nx = (-5\±3)/(8) \n\nx =(-5+3)/(8) \n\nx = (-5-3)/(8) \n\nx = (-2)/(8) \n\nx = (-8)/(8) \n\nx = -(1)/(4) \nx=-1$

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The value of a car is $30,000 and depreciates at a rate of 6.5% each year.What will the value of the car be after 5 years? Round to the nearest cent.

Answers

Answer:$20250

Step-by-step explanation:

6.5% of 30000

6.5/100 x 30000

(6.5 x 30000)/100

195000/100=1950

First year depreciation 30000-1950=28050

Second year=28050-1950=26100

Third year=26100-1950=24150

Fourth year=24150-1950=22200

Fifth year=22200-1950=20250

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