Acombination lock uses 4 numbers, each of which can be 0 to 26. if there are no restrictions on the numbers, how many possible combinations are available?

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.

Assemble the proof by dragging tiles to the statements and reasons column

Solve the inequality and graph 6j - 7 < 5