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Enter the domain in interval notation. To enter , type infinity.Your responseCorrect response(-2,infinity)(-2, infinity)Auto gradedGrade:1/1.0 The vertical asymptote is Your responseCorrect response-2-2Auto gradedGrade:1/1.0 As approaches the vertical asymptote,Your responseCorrect response-infinity-infinityAuto gradedGrade:1/1.0 .As approaches ,Your responseCorrect responseinfinityinfinityAuto gradedGrade:1/1.0 . Which of the following graphs best represents the graph of
?Your responseCorrect responseb=ey=-4.8x=h=he-4.8ln(h) =-4.8g(x) = ln(5x+ 10) + 1.4∞!x=!xg(x)→!x∞g(x)→!g(x)
9/29/20, 5:08 PMSouthern New Hampshire University -Page 15 of 23Auto gradedGrade:1/1.0 Show your work and explain, in your own words, how you arrived at your answers.Answerswith no relevant explanations may receive reduced or no credit.!as x→2,g(x)→ -∞and as x→+∞,g(x)→ ∞
9/29/20, 5:08 PMSouthern New Hampshire University -Page 16 of 23The input must be positive.Subtract .Divide by .The domain of is and the vertical asymptote is .The graph of is shown in the figure below. As , and as , .Question9:Score 0/0Find the exact solution for
If there is no solution, enter NA.Enclose arguments of functions in parentheses and include a multiplication sign between terms. Forexample, * log(h)..5x+ 10 > 05x>-1010x>-25g(x) = ln(5x+ 10) + 1.4(-2,∞)x=-2g(x) = ln(5x+ 10) + 1.4x→2g(x)→ -∞x→+∞g(x)→ ∞-5-50 = 0e2xexc
9/29/20, 5:08 PMSouthern New Hampshire University -Page 17 of 23Your responseCorrect responseln(10)ln(10)Auto gradedGrade:1/1.0 Show your work and explain, in your own words, how you arrived at your answer.Answerswith no relevant explanations may receive reduced or no credit.factor the equation(e^x-10)(e^x+5)=0If any factor on the left side of the equation is equal to 0 the entire expression is now 0 so e^x-10 is equal to 0 and we solve for xx=ln(10)The final solution is all the values that make (ex−10)(ex+5)=0 true.x=ln(10)Ungraded Grade:0/1.0Total grade: 1.0×1/2 + 0.0×1/2 = 50% + 0%Feedback:Factor by the FOIL method.orIf a product is zero, then one factor must be zero.orIsolate the exponentials.Reject the negative equation. Solve the positive equation.Question10:Score 0/0Use the one-to-one property of logarithms to find an exact solution forx=!"-5-50e2xex= 0(-10)(+ 5)exex= 0-10ex= 0+ 5ex= 0ex= 10ex=-5ex= 10x= ln(10)
9/29/20, 5:08 PMSouthern New Hampshire University -Page 18 of 23.If there is no solution, enter NA.The field below accepts a list of numbers or formulas separated by semicolons (e.g.