58 = 3x + y
y = x-2
58 = 3x + x-2
58 = 4x - 2
56 = 4x
14 = x
x-2 = y
y = 16
Let x= the lesser integer=14
Let x+2 = the larger integer=16
3(x)+(x+2)=58
3x+x+2=58
4x+2=58
4x=56
x=14
x+2
(14)+2
16
Let the smaller number be n, then the larger number is n + 2.
3n + (n + 2) = 58 : 4n + 2 = 58 : 4n = 56 : n = 14. then n + 2 = 16
The two consecutive even numbers are 14 and 16.
Suppose x is the smaller integer.
Then the larger one is x + 2
Also, 3*x + (x+2) = 58
ie 4x + 2 = 58
that is 4x = 56
or x = 14
Thus, the two integers are 14 and 16.
The quotient will be less than one.
A square root of a number is a lesser number that when multiplied by it's self equals the larger number. The square root of 16 is 4. 4*4=16
-39 is greater than -259 because the futher up the negatives the greater the number, and the futher down the negatives the lesser the number. With negatives how large the number is, is opposite to positives. With negative numbers the closer to 0 the greater the number, and the futher away from 0 the lesser the number.
Call the integer square roots of the specified pair of numbers l and g for lesser and greater respectively. Then, from the problem statement, g2 - l2 = 105. Possible values for l2 are successively 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, .... , and corresponding values for g2 are 106, 109, 114, 121, ... etc. The first of the latter series that is a perfect square number is 121. Therefore, the square numbers are 121 and 16.
1 km = 1,000 m So 4 km = 4,000 m, and neither is greater.
Let the smaller integer be x, then then larger integer is x + 2, and: 3x + (x + 2) = 58 → 4x = 56 → x = 14 → The two integers are 14 and 16.
The integers are the sqare root of 121 and 81 to the 2/4 power.
Call the unknown higher integer h. Then, from the problem statement, 4h + (h-2) = 98; 5h = 100; h = 20. Thus the two integers are 18 and 20.
Let the two integers be m-1 & m, so the lesser integer is m-1, and the greater integer is m:(m-1) + 3*m = 43 --> 4*m -1 = 43 --> 4*m = 44, m = 11 & m-1 = 10, so 10 & 11Check: 11 times 3 is 33, add 10 is 43
3(n + 2) = n - 10 3n + 6 = n - 10 2n + 6 = -10 2n = -16 n = -8 -8 and -6
6 and 8
Let the two odd numbers be n and n+2, then : 2(n + 2) + 13 = 3n 2n + 4 + 13 = 3n n = 17 The smaller of the two odd numbers is 17 and the larger number is 19.
Let's denote the unknown integer as "x". So now we have two integers, "x" and "4x" because one integer is 4 times the other. So the sum of x+4x= 5x 5x = 5 So x=1
The two integers you are looking for would be 10 and 12. 12X3=36 therefore 10+(12X3)=46. Voila.
Not exactly. All positive integers are greater than all negative integers, but -1 is greater than -7 even though -7 is farther from zero. It's better to think of a number line. Things get greater as you move to the right, lesser as you move to the left.
The quotient will be less than one.
The representation of an odd integer is 2n + 1. So the other consecutive odd integer will be 2n + 3,[(2n + 1) + 2]. Thus, we have:2(2n + 3) + 13 = 3(2n + 1)4n + 6 + 13 = 6n + 3 subtract 4n and 3 to both sides16 = 2n divide by 2 to both sides8 = n2n + 1 = 2(8) + 1 = 16 + 1 = 172n + 3 = 2(8) + 3 = 16 + 3 = 19Therefore, theses numbers are 17 an 19.Check:3(17) - 2(19) = 13 ?51 - 38 = 13?13 = 13 True