Amanufacturing operations consists of 12 operations. however, five of the 12 machining operations must be completed before any of the remaining operations can begin. within each of these two sets, operations can be completed in any order. how many different production sequences are possible?

604800 possibilities

Step-by-step explanation:

There are two sets.

The first set is for the first five machining operations. This set has the following operations: A,B,C,D,E

The first operation to be made can be any of them(A,B,C,D,E). The second will be any of the four remaining ones, the third will be any of the three remaining and so on...

So for the first set there are 5*4*3*2*1 = 5! = 120 possibilities.

Now for the second set, the same logic is applied.

There will be 7*6*5*4*3*2*1 = 7! = 5040 possibilities.

Now taking into account all the 12 operations, separated by set, there will be 120*5040 = 604800 possibilities.

You can understand this calculus as that for a simple possibility of the first set, there are 5040 possibilities of the second set. The are 120 possiblities in the first set, so we have the multiplication in the paragraph above.

the length of o'p' is .

step-by-step explanation:

it is given that the quadrilateral m'n'o'p' is the image of quadrilateral mnop after dilation.

the center of dilation is origin and the scale factor is 1.5.

in dilation the image and preimge are similar figures. it means the sides of image and preimage are proportional.

the scale factor is defined as

the scale factor is 1.5.

multiply both sides by op.

note: if the value of op is given, then multiply the value of op with 1.5 to get the length of side o'p'.

therefore length of o'p' is .