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4n -14 = 22 4n = 22+14 4n = 36 n = 9
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22
Let the number be n & n+1 ; consecutive Their product (multiplication) is n(n+1) = 182 n^(2) + n = 182 n^(2) + n - 182 = 0 Factor ( n +14)(n - 13) = 0 Hence the numbers are 13, & 14.
If the number is n, and the product is of the numbers p and q, the expression is p*q + (n + 15) : the parentheses are not necessary.
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Is wa adde n 1951.
You can solve this in two ways.1) Trial and error. That is, try multiplying two consecutive integers; if the product is too large, try smaller integers; if the product is too small, try larger consecutive integers. 2) Call the two consecutive integers "n" and "n+1", and solve the equation: n(n+1)=210
Call the smaller of the two consecutive integers n. Then, from the problem statement: n(n+2) = 168, or n2 + 2n - 168 = 0, or (n + 14)(n - 12) = 0, which is true when n = -14 or +12. Therefore, the two integers sought are 12 and 14.
One way to find out is write a formula. Let N and N+1 be the two integers, then N(N+1) = 182 N^2 + N - 182 = 0 This is a quadratic equation. If the factors are not obvious, (N -13)(N + 14) , then use the quadratic formula to find N. The factors tell you there are two possible solutions for N; 13 and -14. Now add 1 to these to get the two consecutive integers. 13 & 14 will work and -14 & -13 will work.
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Let the number be n. Then 3n + 2 = -4 3n = -6 n = -3