Let n be the larger integer and n-2 be the smaller integer.
n² - (n - 2)²
= n² - (n - 2)(n - 2)
= n² - (n² - 4n - 4)
= n² - n² + 4n + 4
= 4n + 4
= 4(n + 1)
No matter what value n is, it will always be a multiple of 4.
X+X+3xX+4 divided by X = F
The equation computed shows that the numbers will be 9, 11 and 13.
<h3>How to illustrate the information?</h3>
Let the numbers be:
First number = x
Second number = x + 2,
Third number = x + 4.
Total of the numbers = 33
The equation will be:
= first + second + third number
x + x + 2 + x + 4 = 33
3x + 6 = 33
Collect like terms
3x = 33 - 6
3x = 27.
Divide both side by 3
3x/3 = 27/3
x = 9
First number = x = 9
Second number = x + 2 = 9 + 2 = 11
Third number = x + 4. = 9 + 4 = 13
The numbers will be 9, 11, and 13.
Learn more about equations on:
brainly.com/question/2972832
#SPJ1
Illustrate the equation to find three consecutive odd integers whose sum is 33.
Answer:
Step-by-step explanation:
Graph: f(x) = 1 - ex
You meant f(x) = 1 - e^x, where ^ represents exponentiation.
First, graph the parent function y = e^x. The y-intercept of this graph is (0, 1), and the x-axis is the horizontal intercept. The graph begins on the left in Quadrant 2 and continues upward in Quadrant 1.
Next, reflect this graph in the x-axis, due to the - sign in f(x) = 1 - e^x. The y-intercept is (0, -1) and the horizontal asymptote is the x-axis.
Finally, translate the whole graph upward by 1 unit.
In the first table, y = 4x. So y is proportional to x.
In the second table, y is not proportional to x.
In the third table, 2y = 3x. So, y is proportional to x.
