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Q: How do you make 55 with numbers 1 2 3 4?

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5 x 11 = 55

1 and 54 - added together.

For whole numbers: 1, 2, 5, 10, 11, 55, 110

330

55----------------------------------------------------------------------------------------------From Rafaelrz.You can make, 5! = 120, five digit numbers using 1,2,3,4 and 6.

It is 10*(10+1)/2 = 55

1 2 5 11 22 55 110

There are 11C2 = 11*10/(2*1) = 55 combinations.

11

1, 2, 5, 10, 11, 22, 55, 110

The sum_of_a_series = 1/2 x number_in_series x (first_number + last_number) The first odd number is 2 x 1 - 1 = 1, the 55th odd number is 2 x 55 - 1 = 109 Sum = 1/2 x 55 x (1 + 109) = 1/2 x 55 x 110 = 552 = 3025 It can be seen that the sum of the first n consecutive odd numbers is always n2: sum = n/2(1 + (2n - 1)) = n/2(2n) = n2

(-1)+(-2)=(-3) (-1)(-2)=2

1 and 1/2 + 1 and 1/2

4 *2/2*1

The first 11 Fibonacci numbers are 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, and 55.

There are 56C5 = 56*55*54*53*51/(5*4*3*2*1) = 3,819,816 combinations.

The prime numbers from 1 - 55 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, and 53.

1 1 2 3 5 8 13 21 34 55

The GCF is.... 1 !

1, 2, 5, 10, 11, 22, 55, 110 | 1 x 110, 2 x 55, 5 x 22, 10 x 11.

1, 2, 5, 10, 11, 22, 55, 110.

55 is an odd number. Odd numbers end in 1, 3, 5, 7, or 9. Even numbers end in 0, 2, 4, 6, or 8.

256!

The two numbers are 125 and 55 You can make 2 equations 1) x + y = 180 2) x+ 40 =3y If you rearrange equation 2 you get x=3y-40 You can now put this in for x in equation 1 (3y-40) +y = 180 3y + y = 180 + 40 4y = 220 y=55 If you plug this in for y in equation 1 you get X + 55 = 180 x= 125 Check with equation 2 x+ 40 =3y 125+40=3x55 165=165 Therefore the two numbers are 125 and 55

4 + 1 - 2 = 3