Math Question and Answers


Question and Answers Forum

All Questions Topic List
Vector Questions
Previous in All Question Next in All Question
Previous in Vector Next in Vector
Question Number 159748 by 07814040830 last updated on 20/Nov/21

	Answered by 07814040830 last updated on 20/Nov/21


	Answered by TheSupreme last updated on 21/Nov/21

	${(4 t - 0)}^{2} + {(2 t)}^{2} + {(2 t - 12)}^{2}$ $\frac{\partial d}{\partial t} = 0 \to 2 \times 4 \times (4 t) + 2 \times 2 (2 t) + 2 \times 2 (2 t - 12) = 48 t - 48$ $t = 1$ $4^{2} + 2^{2} + 100 = \sqrt{120}$ ${(5 + 3 t)}^{2} + {(5 + 4 t)}^{2} + {(- 3 - 5 t)}^{2}$ $2 (5 + 3 t) 3 + 2 (5 + 4 t) 4 + 2 (3 + 5 t) 5 = 0$ $30 + 18 t + 40 + 32 t + 30 + 50 t = 0$ $100 + 81 t = 0$ $t = 1$ $8^{2} + 9^{2} + 8^{2} = 128 + 81 = \sqrt{209}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum