2nd time asking, please help. Question in attached picture

Anybody know the answer I need help. Please

Evgen [1.6K]

Answer:

$\dfrac{p^2+9p+2}{p^2-49}$

Step-by-step explanation:

Problems like this require that you recognize that the denominator of the right term is a factor of the denominator of the left term. That is, you're supposed to know how to recognize and factor the difference of two squares.

$\dfrac{5-p}{49-p^2}+\dfrac{p+1}{p-7}=\dfrac{p-5}{p^2-49}+\dfrac{p+1}{p-7}\\\\=\dfrac{p-5}{(p-7)(p+7)}+\dfrac{p+1}{p-7}\cdot\dfrac{p+7}{p+7}\\\\=\dfrac{(p-5)+(p+1)(p+7)}{(p-7)(p+7)}=\dfrac{p-5+p^2+8p+7}{p^2-49}=\dfrac{p^2+9p+2}{p^2-49}$

8 0

3 years ago

Use the given information to write the equation of the parabola.

m_a_m_a [10]

Answer:

x² = -2y

Step-by-step explanation:

The focus is p away from the vertex, and so is the directrix.

To find the equation of the parabola, we must first determine if the parabola is horizontal or vertical.

Horizontal parabola [Standard form]: (y – k)² = 4p(x – h)
Vertical parabola [Standard form]: (x – h)² = 4p(y – k)

If the parabola is vertical, the directrix, and focus will have the same x value but different y value compared to the vertex (h, k). You can also tell if the directrix in in the form y = k – p, and if the focus is in the form (h, k + p).

Likewise, if the parabola is horizontal, the directrix, and focus will have the same y value but different x value compared to the vertex (h,k) . You can also tell if the directrix is in the form x = h – p, and if the focus is in the form (h + p, k).

For this problem, given that the vertex is at the origin (0,0), and that the focus is at the point (0, -½).

You can tell that the x value is the same for the vertex, and focus so this must be a vertical parabola. Because this is a vertical parabola, we can use the form mentioned as earlier (x – h)² = 4p(y – k).

If h = 0, and k = 0, the p value must be the difference between the k of the vertex, and the k of the focus: -½ - 0 → -½.

Now we can just plug in our known information to derive the equation!

h = 0, k = 0, p = -½ → (x - h)² = 4p(y - k) →

(x - 0)² = 4(-½)(y - 0) → x² = -2(y - 0) →

x² = -2y.

Also p = 1/4a, if you are wondering.

So because this is a vertical parabola, x² = -2y is generally the same as y = -1/2x² in standard quadratic form. I just like to think of the horizontal parabola as an inverse quadratic because it is like reflecting over the line y = x.

8 0

3 years ago

Read 2 more answers

What are the solutions to the quadratic equation (5y + 6)2 = 24?

Ivenika [448]

Answer:

y = $\frac{2\sqrt{6} }{5} -\frac{6}{5}$ OR y = $\frac{-2\sqrt{6} }{5} -\frac{6}{5}$

Step-by-step explanation:

Our quadratic equation is: (5y + 6)² = 24.

The first step is to square root both sides:

5y + 6 = ±√24 = ±2√6

Now subtract 6 from both sides:

5y = ±2√6 - 6

Finally divide by 5 from both sides:

y = $\frac{2\sqrt{6} }{5} -\frac{6}{5}$ OR y = $\frac{-2\sqrt{6} }{5} -\frac{6}{5}$

And, those are the solutions to the equation.

7 0

4 years ago

Read 2 more answers

An object in a simulation accelerates

notsponge [240]

Observe the graph below. This graph represents the scenario.

The question is ill formated, the complete question is

In a simulation, a moving object accelerates from rest to 4 meters per second in 2 seconds. For the following three seconds, it increases linearly until it reaches a speed of 10 meters per second. Following three seconds at that speed (acceleration = 0), the item progressively decelerates until it comes to rest two seconds later. Draw the graph of this scenario for 10 seconds?

I'll describe how the graph may show.

It will move diagonally upward from time 0 to 2 seconds until it reaches the y axis at a speed of 4 m/s.

Then, from 2 to 5, the position will move up diagonally until it reaches the y axis at a speed of 10 m/s.

The next 5 to 8 seconds will be horizontal.

After that, it will descend diagonally.

Observe the graph below. This graph represents the scenario.

Learn more about Acceleration here-

brainly.com/question/21509870

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