Linear graph problem in the picture ​

Find the arc length parameter along the given curve from the point where tequals=0 by evaluating the integral s(t)equals=integral from 0 to t startabsolutevalue bold v left parenthesis tau right parenthesis endabsolutevalue d tau∫0tv(τ) dτ. then find the length of the indicated portion of the curve r(t)equals=1010cosine tcost iplus+1010sine tsint jplus+88t k, where 0less than or equals≤tless than or equals≤startfraction pi over 3 endfraction π 3.

Use the expression below.–4b + 8c + 12 – 8b – 2c + 6part asimplify the expression. enter your answers in the boxes. b + c + part bfactor the simplified expression using the gcf. a. 2(–2b + c + 3) b. 3(–2b + c + 3) c. 4(–2b + c + 3) d. 6(–2b + c + 3)part cwhat is the value of the expression when b = 2 and c = –3? enter your answer in the box.

The larger of two number is 4 less than twice the smaller number. the sum is 41. find the numbers