Answer:
(x+5/2)^2 = 0
Step-by-step explanation:
4 x^{2} +20x +25 =0
Divide by 4
4/4 x^{2} +20/4x +25/4 =0
x^2 +5x +25/4 =0
Subtract 25/4 from each side
x^2 +5x +25/4 -25/4 =-25/4
x^2 +5x =-25/4
Take the coefficent of x
5
Divide by 2
5/2
Square it
25/4
Add it to each side
x^2 +5x +25/4 =-25/4+25/4
(x+5/2)^2 = 0
Take the square root of each side
x+5/2 = 0
x = -5/2
Answer:
-10
Step-by-step explanation:
well you know that
so we have
<span>1. We analyze the limit by approaching it from both the left and the right.
From the left: f(x) = x + 10 (for x < 8), as x --> 8, f(x) --> 18
From the right: f(x) = 10 - x (for x >= 8), as x --> 8, f(x) --> 2
Since the limits on either side do not converge to the same point, the limit does not exist (this is choice C).
2. </span>Using a similar approaching as in #1:
<span><span>From the left: f(x) = 5 - x (for x < 5), as x --> 5, f(x) --> 0
At x = 5 itself: f(x) = 8
From the right: f(x) = x + 3 (for x > 5), as x --> 5, f(x) --> 8</span>
Although the value at x = 5 matches with the limit when approaching from the right, the limit when approaching from the left doesn't match, so the limit does not exist (choice D).
3. </span><span><span>From the left: f(x) = 5x - 9 (for x < 0), as x --> 0, f(x) --> -9
From the right: f(x) = |2 - x| (for x >= 0), as x --> 0, f(x) --> 2
</span>Again, since the limits when approaching from the left and right don't match, the limit does not exist. (This is Choice D).
4. lim 1/(x - 4) as x -->4-
If we are approaching x = 4 from the left, we can test values such as 3, 3.9, 3.99, 3.999, approaching 4. For x = 3, f(x) = -1. For x = 3.9, f(x) = -10. For x = 3.99, f(x) = -100. For x = 3.999, f(x) = -1000. This shows that the value continues to go towards negative infinity.
If we were to graph these 4 points on the Cartesian plane, it would also show a curve to slopes downwards to negative infinity, with the vertical asymptote at x = 4. The correct answer is Choice C) -∞ ; x = 4.
5. </span>f(x) = (x+1)(x-1) / [(x+1)(x-2)] is an example of a function with both a removable and non-removable discontinuity.
In this case, because x+1 cancels out from the numerator and denominator, it results in a hollow or missing point (removable) discontinuity at x = -1. This means that the limit still exists as x --> -1. On the other hand, x = 2 is a non-removable discontinuity, since it cannot be cancelled out, and it will be an asymptote.
Answer:
The number of persons using the elevator at any hour is never going to be less that 15.
Step-by-step explanation:
To solves this you have to suppose that there are at least 15 persons on the elevator, and the equation is converted into an inequation:
Now you transform the inequation back to an equation to solve it:
You need to know if there is any negative solution for the equation, to do this you can use the discriminant for a quadratic equation:
In this case, you have a=1, b=-10, c=25
Since the discriminant is 0 and a<0 the equation always is going to be positive. Therefore, the number of persons using the elevator at any hour is never going to be less than 15.
