Let → r ( t ) = < − 5 t 4 + 1 , t 5 − 2 t 3 , 3 ln ( 3 t ) > Find a parametric equation of the line tangent to → r ( t ) at the point ( − 404 , 189 , 6.592 ) This problem has been solved! You'll get a detailed solution from a subject matter expert that helps you learn core concepts.
The position vector of a particle is r (t). Find the requested vector. Find the requested vector. 5) The velocity at t = 0 for r ( t ) = ln ( t 3 − 7 t 2 + 4 ) i − 2 t + 9 j − 5 cos ( t ) k A) v ( 0 ) = 4 1 i − 3 1 j B) v ( 0 ) = 3 1 j C) v ( 0 ) = − 3 1 j D) v ( 0 ) = 4 1 i − 6 1 j + 5 k
Determine whether S is a basis for P 3. S = {4t - t 2 , 5 + t 3 , 3t + 5, 2t 3 - 3t 2 } Thank you, please show work. There are 2 steps to solve this one.
What are the solutions to the equation 2t^5+2t^4-144t^3=0 ? The solutions to the equation 2t^5+2t^4-144t^3=0 are t=0,t=8,t=-9 Find the zeros of 2t^5+2t^4-144t^3=0
Explanation: just simply multiply 5 with 2t which will give us 10t and then multiply 5 with 3 this will give us 15. Answer link. 10t + 15 just simply multiply 5 with 2t which will give us 10t and then multiply 5 with 3 this will give us 15.
Click here 👆 to get an answer to your question ️ 5(2t-3)-7.5t=0.5(12-t) See what teachers have to say about Brainly's new learning tools!
Basic Math. Solve for t 3/5t-1/10t=t-5/2. 3 5 t − 1 10 t = t − 5 2 3 5 t - 1 10 t = t - 5 2. Simplify 3 5t− 1 10t 3 5 t - 1 10 t. Tap for more steps t 2 = t− 5 2 t 2 = t - 5 2. Move all terms containing t t to the left side of the equation. Tap for more steps − t 2 = − 5 2 - t 2 = - 5 2.
How do you solve 2(v + 2) = 3v + 3 ? v =1 Explanation: 2(v +2) =3v+3 First distribute the terms on the left side: 2v +4 = 3v +3 Isolate v by 4 (h+2)=3 (h-2) One solution was found : h = -14 Rearrange: Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
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