Acustomer ordered fourteen zingers. zingers are placed in packages of four, three, or one. in how many different ways can this order be filled?

fourteen zingers. zingers are placed in packages of four, three, or one

4*3=12

1*2=2

4*2=8

3*2=6

4*2=8

1*6=6

4*2=8

3*1=3

1*3=3

4*1=4

3*3=9

1*1=1

4*1=4

3*2=6

1*4=4

4*1=4

3*1=3

1*7=7

4*1=4

1*10=10

3*4=12

1*2=2

3*3=9

1*5=5

3*2=6

1*8=8

3*1=3

1*11=11

1*14=14

13 ways

The number of different ways this order can be filled is:

                              13 ways.

A customer ordered fourteen zingers.

zingers are placed in packages of​ four, three, or one.

Case-1

If the number of packages of four are: 3

Then 4×3=12 zingers.

a)

So, there will be zero packet of 3.

and 2 packs of 1.

Since, 12+2×1=14

Case-2

If the number of packages of four are: 2

i.e. 4×2= 8 zingers.

a)

If number of packages of 3 are: 2

i.e. 3×2=6

then number of package of 1 have to be 0.

b)

If number of packages of 3 are: 1

i.e. 3×1=3 zingers

Number of packages of 1 zinger will be: 3

i.e. 8+3+3=14

c)

If number of packages of 3 are: 0

Then number of packets of 1 will be:  6

Case-3

If the number of packages of four are: 1

a)

If number of packages of 3 are: 3

i.e. 3×3=9 zingers.

Hence, number of packets of 1 have to be: 1

b)

If number of packages of 3 are: 2

i.e. 3×2=6 zingers

Then number of packets of 1 have to be: 4

c)

If number of packages of 3 are: 1

i.e. 3×1=3 zingers

Then number of packets of 1 have to be: 7

d)

If number of packages of 3 are: 0

i.e. 3×0=0 zingers

Then number of packets of 1 have to be: 10

Case-4

If the number of packages of four are: 0

a)

If number of packages of 3 are: 4

i.e. 3×4=12 zingers

Then number of packets of 1 have to be: 2

b)

If number of packages of 3 are: 3

i.e. 3×3=9 zingers

Then number of packets of 1 have to be: 5

c)

If number of packages of 3 are: 2

i.e. 3×2=6 zingers

Then number of packets of 1 have to be: 8

d)

If number of packages of 3 are: 1

i.e. 3×1=3 zingers

Then number of packets of 1 have to be: 11

e)

If number of packages of 3 are: 0

i.e. 3×0=0 zingers

Then number of packets of 1 have to be: 14

There are total 13 ways of doing so.

( 1 from case-1

3 from case-2

4 from case-3

and 5 from case-4

i.e.  1+3+4+5=13 ways )

the answer is c: divide the arc's degree measure by 360°, then multiply by the circumference of the circle.

system of linear equation

step-by-step explanation:

as given in the data

a= 20 , b= -12

let us suppose that it is a straight line equation

now we know the formula for the system of linear equation

b= ma + c )

here m is slope of a straight line which is equal to formula

m=   (ii)

and here c is a constant

now putting the value of a and b in equation (ii)

m=

m=

putting the value of a,b and m in equation (i) gives us

-12 = + c

-12=-12 + c

adding 12 to both sides

-12 +12 = -12 +12 +c

0 =c

or c=0

as the value of the constant is zero which gives us that it is a linear equation and this order pair satisfies the system of linear equations.