Entered (2e^2cos(t)-e^(2t)sin(t))i+ (2e^(2t)sin(t)+e^(2t)j+2e^(2t)k 3*[e^(2*t)] (((-2sin(t)-cos(t))/3)i+2cos(t)- sin(t))

Business, Finance, Economics, Accounting, Operations Management, Computer Science, Electrical Engineering, Mechanical Engineering, Civil Engineering, Chemical Engineering, Algebra, Precalculus, Statistics and Probabilty, Advanced Math, Physics, Chemistry, Biology, Nursing, Psychology, Certifications, Tests, Prep, and more.
Post Reply
answerhappygod
Site Admin
Posts: 899603
Joined: Mon Aug 02, 2021 8:13 am

Entered (2e^2cos(t)-e^(2t)sin(t))i+ (2e^(2t)sin(t)+e^(2t)j+2e^(2t)k 3*[e^(2*t)] (((-2sin(t)-cos(t))/3)i+2cos(t)- sin(t))

Post by answerhappygod »

Entered 2e 2cos T E 2t Sin T I 2e 2t Sin T E 2t J 2e 2t K 3 E 2 T 2sin T Cos T 3 I 2cos T Sin T 1
Entered 2e 2cos T E 2t Sin T I 2e 2t Sin T E 2t J 2e 2t K 3 E 2 T 2sin T Cos T 3 I 2cos T Sin T 1 (94.82 KiB) Viewed 12 times
Entered (2e^2cos(t)-e^(2t)sin(t))i+ (2e^(2t)sin(t)+e^(2t)j+2e^(2t)k 3*[e^(2*t)] (((-2sin(t)-cos(t))/3)i+2cos(t)- sin(t))/3)j +0k ([-2*sin(t)-cos(t)]/[sqrt(5)])*i+ ([2*cos(t)- sin(t)]/[sqrt(5)])*j+0*k Answer Preview -2 sin(t) – cos(t) √5 3e²t ∙i+ 2 cos(t) = sin(t) √5 -j+0k Result incorrect correct incorrect correct Message Operands for '+' must be of the same type Operands for '+' must be of the same type
Join a community of subject matter experts. Register for FREE to view solutions, replies, and use search function. Request answer by replying!
Post Reply