What a delightful little problem ! (Partly because I could see
right away how to do it, and had the answer in a few minutes,
after a lot of impressive-looking algebra on my scratch-paper.)
Three consecutive integers are . . . x, x+1, and x+2
The smallest two are . . . x and x+1
Their product is . . . . . x(x+1)
5 times the largest one is . . . 5(x+2)
5 less than that is . . . . . . 5(x+2)-5
Now, the conditions of the problem say that <u>x (x + 1) = 5 (x+2) - 5</u>
THAT's the equation we have to solve, to find 'x' .
Eliminate parentheses: x² + x = 5x + 10 - 5
Combine like terms: x² + x = 5x + 5
Subtract 5x from each side: x² - 4x = 5
Subtract 5 from each side: <u>x² - 4x - 5 = 0</u>
You could solve that by factoring it, or use the quadratic equation.
Factored, it says that (x + 1) (x - 5) = 0
From which <em>x = -1</em>
and <em>x = +5</em>
We only want the positive results, so our three consecutive integers are
5, 6, and 7 .
To answer the question, the smallest one is <em><u>5 </u></em>.
<u>Check</u>:
5 x 6 ? = ? (7 x 5) - 5
30 ? = ? (35) - 5
30 = 30
yay !
<h3>
Answer:</h3>
- a_n = -3a_(n-1); a_1 = 2
- a_n = 2·(-3)^(n-1)
<h3>
Step-by-step explanation:</h3>
A) The problem statement tells you it is a geometric sequence, so you know each term is some multiple of the one before. The first terms of the sequence are given, so you know the first term. The common ratio (the multiplier of interest) is the ratio of the second term to the first (or any term to the one before), -6/2 = -3.
So, the recursive definition is ...
... a_1 = 2
... a_n = -3·a_(n-1)
B) The explicit formula is, in general, ...
... a_n = a_1 · r^(n -1)
where r is the common ratio and a_1 is the first term. Filling in the known values, this is ...
... a_n = 2·(-3)^(n-1)
Answer: 92
Step-by-step explanation: Subtract 257 & 162 to get your answer :)
Answer:
<h2>y = -5x -18</h2>
Step-by-step explanation:
Plug-in values in point -slope form ;
