How many 4-digit multiples of 21 and 15 are there and 4-digit multiples of either 21 or 15 are there?

943

Step-by-step explanation:

21| 1000

   |

       47 | 13

21-13=8

1000+8=1008

21|9999

  |

     476|3

9999-3=9996

1008,1029,...,9996

let n be number of terms

a=1008,d=21,l=9996

9996=1008+(n-1)21

9996-1008=(n-1)21

8988=(n-1)21

n-1=8988/21

n-1=428

n=428+1=429

again 15|1000

            |

                66|10

15-10=5

1000+5=1005

15|9999

  |

      666|9

9999-9=9990

1005,1020,...,9990

a=1005,d=15,l=9990

let n be number of terms.

9990=1005+(n-1)15

9990-1005=(n-1)15

8985=(n-1)15

n-1=8985/15

n-1=599

n=599+1=600

21=3×7

15=3×5

L.C.M.=3×5×7=105

105)1000(9

      -945

         55

105-55=50

1000+50=1050

105)9999(

     -945

         549

       - 525

            24

9999-24=9975

1050,1155,...,9975

let n be number of terms.

d=105

l=a+(n-1)d

9975=1050+(n-1)105

9975-1050=(n-1)105

8025=(n-1)105

n-1=8925/105

n-1=85

n=85+1=86

so there are 86 multiples of 21 and 15

(A∪B)=A+B-(A∩B)

multiples of 21 or 15=429+600-86=1029-86=943

(23*2)-12=x     x=34

step-by-step explanation:

the polynomial is a perfect square trinomial of the form a^2-2ab+b^2:

option a. true

step-by-step explanation:

9x^2-12x+4

comparing with the form a^2-2ab+b^2

1.) a^2=9x^2

solving for a: square root both sides of the equation:

sqrt(a^2)=sqrt(9x^2)

a=sqrt(9) sqrt(x^2)

a=3x

2.) b^2=4

solving for b: square root both sides of the equation:

sqrt(b^2)=sqrt(4)

b=2

to be a perfet square trinomial 2ab must be equal to 12x. let's check:

2ab = 2 (3x) (2)

2ab = 12x

correct