P2 - 5p = 0 solve the equation

Answer:

Step-by-step explanation:

Add 29 to both sides of this equation, obtaining:  x² - 10x + 25.  At this point it becomes obvious that this is a perfect square, the square of x - 5.

Thus, in the first two blanks, write x - 5.

In the second two blanks, write 5 (since 5 is the root corresponding to the factor x - 5).

Step-by-step explanation:

To make the function f(x) = {sin(1/x), x ≠ 0; k, x = 0} continuous at x = 0, we need to find the value of k that ensures the limit of f(x) as x approaches 0 exists and is equal to k.

First, let's find the limit of sin(1/x) as x approaches 0:

lim(x -> 0) sin(1/x)

This limit does not exist because sin(1/x) oscillates wildly as x gets closer to 0. Therefore, in order for the function to be continuous at x = 0, we need to choose k such that it compensates for the oscillations of sin(1/x) as x approaches 0.

A suitable choice for k is 0 because the limit of sin(1/x) as x approaches 0 is undefined, and setting k = 0 ensures that f(x) becomes a continuous function at x = 0.

So, the correct choice is:

d. None (k = 0)

Final answer:

The value of k that would make the function f(x) = sin(1/x) when x ≠0 and f(x) = k when x=0 continuous at x=0 doesn't exist. This is because the limit of sin(1/x) as x approaches 0 is undefined, hence the function cannot be made continuous at x = 0 for any value of k.

Explanation:

To find the value of k that makes the function continuous at x=0, we can apply the definition of continuity, which states that a function, f(x), is continuous at a certain point, x0, if three conditions are met:

• the function is defined at x0
• the limit as x approaches x0 of f(x) exists
• the limit as x approaches x0 of f(x) is equal to f(x0)

In the case of the function f(x) = sin(1/x), the value for x = 0 is undefined, but we've been given that f(0) = k. To make the function continuous at x = 0, the value of k should ideally be equal to the limit of sin(1/x) as x approaches 0.

However, as x approaches 0, sin(1/x) oscillates between -1 and 1, making the limit non-existent. Because the limit does not exist, the function is not continuous at x=0 no matter the chosen value of k. Therefore, the correct answer is (d) None.

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Answer:

Option D

Step-by-step explanation:

Vertices of the polygon are A(1, 3), B(7, 3), C(7, 7), D(4, 7)

Measure of AB = 7 - 1 = 6 units

Measure of BC = 7 - 3 = 4 units

Measure of CD = 7 - 4 = 3 units

From the figure attached,

Area of the trapezoid = [Here, AB and CD are the parallel bases and BC is the height of the trapezoid]

                                    =

                                    = 18 square units

Therefore, Option D will be the correct option.

The correct answers are:

Question 1 - Part A: f(t)=(t+3)²-29; Part B: (-3, -29), minimum; Question 2 - Part A: H(1) = 124, g(1) = 76.8; H(2) = 166, g(2) = 125.6; H(3) = 176, g(3) = 174.4; H(4) = 154, g(4) = 223.2; Part B: Between 3 and 4 seconds, because that is where the values of g(t) catch up with H(t).

Explanation:

Our quadratic function is in the form f(x)=ax²+bx+c. Our value of a is 1, b is 6, and c is -20.

To write a quadratic in vertex form, first take half of the b value and square it: (6/2)² = 3² = 9. This is what we will add and subtract to the function:

f(t) = t²+6t+9-20-9

The squared portion will be (t+b/2)²:

f(t) = (t+3)²-20-9

f(t) = (t+3)²-29

Vertex form is f(x) = a(x-h)²+k, where (h, k) is the vertex; in our function, (h, k) is (-3, -29).

Since the value of a was a positive, this parabola opens upward; this makes the vertex a minimum.

For Question 2 Part A, substitute the values 1, 2, 3 and 4 in H(t) and g(t).

For Part B, we can see that the values of g(t) are much less than that of H(t) until 3 seconds. From there, we can see that g(t) passes H(t). This means that the solution point, where they intersect, is between 3 and 4 seconds.