Compute the permutations and combinations. how many two-digit, positive integers can be formed from the digits 1, 3, 5, and 9, if no digit is repeated?

The number of two-digit, positive integers that can be formed from the digits 1, 3, 5, and 9, if no digit is repeated are:

                                12

We are asked to find the number of two-digit number that can be formed using the digits:

   1,3 ,5 and 9

We know that the number of such digits possible are:

    4×3=12

( Since at the first place there are 4 choices as any of the 4 numbers could come at the first place and also at the second place there are a choice of 3 numbers as the digit's can't be repeated )

       Hence, the answer is:

                    12

10a^3+277a^2-132a-792

step-by-step explanation:

(2a+11)(5a-12)(foil)

=10a^2-24a+55a-132 =10a^2+31a-132

(10a^2+31a-132)(a+6)

=10a^3+60a^2+31a^2+186a^2-132a-792 =10a^3+277a^2-132a-792

i hope this is what you are looking for : )

37/3

step-by-step explanation:

step 1: simplify both sides of the equation.

7x+8(x+

1

4

)=3(6x−9)−8

7x+(8)(x)+(8)(

1

4

)=(3)(6x)+(3)(−9)+−8(distribute)

7x+8x+2=18x+−27+−8

(7x+8x)+(2)=(18x)+(−27+−8)(combine like terms)

15x+2=18x+−35

15x+2=18x−35

tep 2: subtract 18x from both sides.

15x+2−18x=18x−35−18x

−3x+2=−35

step 3: subtract 2 from both sides.

−3x+2−2=−35−2

−3x=−37

step 4: divide both sides by -3.

−3x

−3

=

−37

−3

x=

37

3