Mathematics, 19.07.2021 18:10, angeline2004
Apply the square root principle to solve (x – 5)2 – 40 = 0. Question 14 options: A) x = –5 + 2√10 , x = –5 – 2√10 B) x = –5 + 2i√10, x = –5 – 2i √10 C) x = 5 + 2√10 , x = 5 – 2√10 D) x = 5 + 2i√10 , x = 5 – 2i√10
Answers: 2
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Solve 3x-18=2y and 5x-6y=14 by elimination or substitution . show all !
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The pyramid below was dissected by a horizontal plane which shape describes the pyramid horizontal cross section
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The position of a moving particle is given by the position function: f(t)=-9t-t^2-0.2t^3+0.1t^4 0 a. at what time does the particle reverse direction? b. when is the displacement positive? (round one decimal place and answer in interval notation) c. when is the displacement negative? (round one decimal place and answer in interval notation) d. when is the particle’s acceleration positive? (round one decimal place and answer in interval notation) e. when is the particle’s acceleration negative? (round one decimal place and answer in interval notation)
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Apply the square root principle to solve (x – 5)2 – 40 = 0. Question 14 options: A) x = –5 + 2√10 ,...
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