18. 95 = -4(m – 3)² – 5 %3D | Мary 95 = -4(m - 3)² – 5 +5 +5 100%3D - 4(т - 3)° + V100 = -2(m – 3) + 10 = -2(m – 3) | | 10 = -2(m – 3) -10 = - 2(m – 3) - 3) | 10 = m - 3 -2 -10 - — т — 3 -2 | | -5 = m - 3 5%3D т — 3 | -2 = m 8 = m
18. 95 = -4(m – 3)² – 5 %3D | Мary 95 = -4(m - 3)² – 5 +5 +5 100%3D - 4(т - 3)° + V100 = -2(m – 3) + 10 = -2(m – 3) | | 10 = -2(m – 3) -10 = - 2(m – 3) - 3) | 10 = m - 3 -2 -10 - — т — 3 -2 | | -5 = m - 3 5%3D т — 3 | -2 = m 8 = m
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
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(If possible could you please type out the steps? thankyou!)
![18. 95 = -4(m – 3)² – 5
Mary
95 = -4(m - 3)2 - 5
+5
+5
100 = -4(m - 3)²
± V100 = -2(m – 3)
± 10 = -2(m - 3)
|
%3D
10%3D -2(т — 3) -103 -2(т - 3)
10
= m - 3
-2
-10
= m – 3
-2
-5 = m - 3
5 = m – 3
-2= m
8 = m
|](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4a38e74a-5625-47be-a66b-cba773f8abfc%2F12a54ca3-ac5c-42b5-adcd-c03bf3d10872%2Fkgow18y_processed.jpeg&w=3840&q=75)
Transcribed Image Text:18. 95 = -4(m – 3)² – 5
Mary
95 = -4(m - 3)2 - 5
+5
+5
100 = -4(m - 3)²
± V100 = -2(m – 3)
± 10 = -2(m - 3)
|
%3D
10%3D -2(т — 3) -103 -2(т - 3)
10
= m - 3
-2
-10
= m – 3
-2
-5 = m - 3
5 = m – 3
-2= m
8 = m
|
![For Exercises 17 and 18, the students were asked to solve
the given equation using the square root property. Explain
what the student did wrong. Give the correct answer.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4a38e74a-5625-47be-a66b-cba773f8abfc%2F12a54ca3-ac5c-42b5-adcd-c03bf3d10872%2Fa3g8rmg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:For Exercises 17 and 18, the students were asked to solve
the given equation using the square root property. Explain
what the student did wrong. Give the correct answer.
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