MATH 225 Week 1 Test | Questions and Answers | 2026 Update - Grade: 100%

MATH 225 Week 1 Test | Questions and Answers | 2026 Update - Grade: 100%.
Instructions:
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THIS IS A TIMED TEST. YOU HAVE 3 HOURS TO COMPLETE THE TEST ONCE YOU CLICK "START." You can start and stop the Test if you
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need to; however, the time will continue to elapse. You can also skip questions and go back to them as needed during the test. Use the 'skip' bu
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tton to skip a question and question navigation pull-down menu to jump back to any questions you skipped.
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Once you have completed the Test online, click “Submit Answers.” Your answers will be scored and the answer key with step-by-
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step solutions will become available.
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Questions? Reach out to us at . We’re here and happy to help.
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Questions Limits Points Due Date cv




15 Questions cv 180 Minutes cv 100 pts possible cv cv No due date. cv cv




Attempt 1 cv 100% (100 of 100) cv cv cv Completed on 03/09/25 at 01:42AM cv cv cv cv




Score for this quiz: 100% ( 100 /100) Submitted
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Mar 9 at 1:42am
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This attempt took 40 minutes.
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Question 1 : 6.65 ptsSkip to question text. cv cv cv cv cv cv cv




If −x2+x+1≤f(x)≤−x+2 for all x, find limx→1f(x).
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Enter only the value of the limit in the space provided below. If the answer is not an integer, enter it as a fraction in simplest form. D
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o not enter a mixed number. If the limit does not exist, enter “does not exist”.
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Your Answer: 1 cv cv



Correct Answer(s): c v



1
Since −x2+x+1≤ f(x)≤ −x+2, by the Squeeze Theorem, limx→1−x2+x+1≤ limx→1f(x)≤ limx→1−x+2
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−12+1+1≤ limx→1f(x)≤ −1+2 cv cv cv




1≤ limx→1f(x)≤ 1Thus, limx→1f(x)=1.
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6..65 c v c v

, Question 2 : 6.65 ptsSkip to question text. f(x) cv cv cv cv cv cv cv cv




={x, x<1x+2, x≥1Evaluate limx→ 1 + f(x). 3
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2
1
The one-sided limit does not exist.
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The plus-sign in the limit indicates a right-handed limit. For right-
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handed limits, you only have to consider the domain values greater than the one being approached in the limit. f(x)={x,
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x<1x+2, x≥ 1limx→ 1 + f(x)=limx→ 1 + (x+2)=(1+2)=3 cv cv c v cv cv c v cv cv




6..65 c v c v




Question 3 : 6.65 ptsSkip to question text. cv cv cv cv cv cv cv




After jumping out of a plane at t=0, a skydiver's altitude in the air in meters isgiven by the position functionp(t)=−50t2+200t+2350, where t is
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thetime in seconds. At what time will theparachuter be 100 meters from the ground?
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t = 9 or t = −5
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t = 9.28
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t = −477650
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t= 9
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Set p(t)=100 and solve for t.p(t)=−50t2+200t+2350=100−50t2+200t+2250=0−t2+4t+45=0t2−4t−45=0(t−9)(t+5)=0t=9 or t=−5Since t=−5 doesn't ma ke
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sense in thecontext of the problem (the skydiver jumpedout of the plane at t=0), we omit thissolution.
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6..65 c v c v




Question 4 : 6.65 ptsSkip to question text. cv cv cv cv cv cv cv




What is the change in U.S. population from July 1, 1985 to July 1, 1987?
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Nat'l. Po Pop. Chang
Date
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p. e
July 1, 272,690,8
2,442,810
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1999 13
July 1, 270,248,0
2,464,396
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1998 03
July 1, 267,783,6
2,555,035
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1997 07
July 1, 265,228,5
2,425,296
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1996 72
July 1, 262,803,2
2,476,255
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1995 76
July 1, 260,327,0
2,544,413
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1994 21
July 1, 257,782,6
2,752,909
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1993 08