1) Refer back to Lesson 2, Part A. Assume that the average shoulder width of the people in the line was 1.325 feet. How long would the line be if it contained 10 million people? Express your answer in feet.​


Answer 1


13,250,000 feet or 13.25 million feet

Step-by-step explanation:

Each person in the line has a shoulder width of 1.325 feet.

For each person, there is a width of 1.325 ft. There are 10 million people.


10 million = 10,000,000

The total length must be:

1.325 * 10,000,000

1.325 * 10,000,000 = 13,250,000

Answer: 13,250,000 feet or 13.25 million feet

Answer 2

Final answer:

By simply multiplying the average shoulder width of the people (1.325 feet) by the total number of people in the line (10 million), the total length of the line would be 13,250,000 feet.


This problem can be solved by using simple multiplication. Given that the average shoulder width of the people in the line is 1.325 feet, we simply need to multiply this by the total number of people in the line, which is 10 million in this case.

So, 1.325 feet * 10,000,000 = 13,250,000 feet.

Therefore, if there were 10 million people in line, and the average shoulder width of each person is 1.325 feet, the total length of the line would be 13,250,000 feet.

Learn more about Multiplication here:



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Someone please answer!




Step-by-step explanation:

y1= 1


x1= -10


m= slope

m= y2-y1/x2-x1

m=6-1/0 - - 10

m= 6-1/0+10



What’s the correct answer for this?




Step-by-step explanation:

The lines have equal slopes

-6/3= -2


-8/4 = -2 so both slopes are equal hence the lines are parallel


  • The lines have proportional slope

Step-by-step explanation:

because , they determined as slope !

hope it helpful !

mark as brainliest

Hey I need help with my coordinates ​


Ok send the picture so I can help you.


what is the question? I can help

Step-by-step explanation:

Use the product rule to calculate the derivatives of
( ax² + bx + c ) ( cx + d )


\n \sf\longmapsto (d)/(dx)(ax^2+bx+c)(cx+d)

\boxed{\sf (d)/(dx)f(x).g(x)=f(x)(d)/(dx)g(x)+g(x)(d)/(dx)f(x)}

  • c and d are constants

\n \sf\longmapsto (ax^2+bx+c)(d)/(dx)(cx+d)+(cx+d)(d)/(dx)(ax^2+bx+c)

\n \sf\longmapsto (ax^2+bx+c)(c)+(cx+d)(2ax+b)

\n \sf\longmapsto acx^2+bcx+c^2+2acx^2+bcx+2adx+bd

\n \sf\longmapsto 3acx^2+2bcx+2adx+bd+c^2


• Product rule is as below:

{ \boxed{ \tt{ \:  (dy)/(dx) = { \huge{v}} (du)/(dx)  + { \huge{u}} (dv)/(dx)  }}} \n

  • u is (ax² + bx + c)
  • v is (cx + d)
  • du/dx is 2ax + bx
  • dv/dx is c

\hookrightarrow \: { \rm{ (dy)/(dx) = (cx + d)(2ax + b) + (ax {}^(2)  + bx + c)(c) }} \n  \n { \rm{ (dy)/(dx) =  (2ac {x}^(2)  + bcx + 2adx + db) + (ac {x}^(2)  + bcx +  {c}^(2) )}} \n  \n { \boxed{ \rm{ (dy)/(dx)  = 3ac {x}^(2)  +  \{2bcx + 2adx \}x + (db +  {c}^(2)) }}}

A traveler has 8 pieces of luggage. How many ways can he select 4 pieces of luggage for a trip


He can select it in 32 ways


Step-by-step explanation:

Please help me understand I am confused


9514 1404 393


a) The velocity curve is linearly increasing from 0 to 6 m/s over a period of 2 seconds, then linearly decreasing from 6 m/s to 0 over the same period. The acceleration is the rate of change of velocity, so for the first half of the motion the acceleration is a constant (6 m/s)/(2 s) = 3 m/s². Similarly, over the second half of the motion, the acceleration is a constant (-6 m/s)/(2 s) = -3 m/s².

The distance traveled is the integral of the velocity, so the linearly increasing velocity will cause the distance vs. time curve to have a parabolic shape. The shape will likewise be parabolic, but with decreasing slope, as the velocity ramps down to zero. Overall, the distance versus time curve will have an "S" shape.

The motion (position and velocity) will be continuous, but the acceleration will not be. There will be a significant "j.erk" at the 2-second mark where acceleration abruptly changes from increasing the velocity to braking (decreasing the velocity).


b) The attachment shows the (given) velocity curve in meters per second and its integral, the position curve, in meters.

The integral in the attached works nicely for machine evaluation. For hand evaluation, it is perhaps best written piecewise:

  s(t)=\begin{cases}\displaystyle\int_0^t{3x}\,dx\qquad\text{for $x\le2$}\n\n\displaystyle6+\int_2^t{(12-3x)}\,dx\qquad\text{for $2<x\le4$}\end{cases}