Good afternoon,
x= first odd integer
x+2 = second odd integer
x+4= third intenger
The sum of the trhree integer is 195, so:
x + (x+2) + (x+4) = 195
3x = 195 - 4 - 2
3x=189
x= 63
Then the Answer is: 63, 65 and 67
Answer:
The roots are;
x = (2 + i)/5 or (2-i)/5
where the term i is the complex number representing the square root of -1
Step-by-step explanation:
Here, we want to use the completing the square method to solve the quadratic equation;
f(x) = -5x^2 + 4x -1
Set the function to zero
0 = -5x^2 + 4x - 1
So;
-5x^2 + 4x = 1
divide through by the coefficient of x which is -5
x^2 - 4/5x = -1/5
We now take half of the coefficient of x and square it
= -2/5^2 = 4/25
add it to both sides
x^2 - 4x/5 + 4/25= -1/5 + 4/25
(x- 2/5)^2 = -1/5 + 4/25
(x - 2/5)^2 = -1/25
Take the square root of both sides
x - 2/5 = √( -1/25
x - 2/5 = +i/5 or -i/5
x = 2/5 + i/5 or 2/5 - i/5
If the degree of numerator and denominator are equal, then limit will be leading coefficient of numerator divided by the
leading coefficient of denominator.
So then the limit would be 3/1 =
3.
Alternatively,

Hope this helps.
The scenario can be described using a piecewise function like:
f(x) = 1/x if x < c.
f(x) = x if x = c
f(x) = 1/(x + 73) if x > c.
<h3>
When the value exists but the limit does not?</h3>
Remember that the limit only exists if the limit from left and the limit from the right give the same value.
Then, we can just define a piecewise function of the form:
f(x) = 1/x if x < c.
f(x) = x if x = c
f(x) = 1/(x + 73) if x > c.
Clearly, this is not a continuous function.
Notice that:

So the limits from left and right are different, then:

Does not exist.
If you want to learn more about limits:
brainly.com/question/5313449
#SPJ1
1) 94
2) 77
3) 290
4) 682
5) 45
6) 925
7) 416