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3 years ago

Which graph represents the function of f(x) = 9x 2 -36/3x+6 Please help

Mathematics

1 answer:

UkoKoshka [18]3 years ago

6 0

The function is:

         (9x^2 - 36)
f(x) = ---------------------
                3x + 6

You can simplify that function to a linear function for all x for which 3x + 6 ≠ 0

=> x ≠ - 2.

So, for x ≠ - 2, you can do:

         (9x^2 - 36)
f(x) = --------------------- =
               3x + 6

          9(x^2 - 4)
f(x) = ---------------------
                3(x + 2)

         3(x + 2)(x - 2)
f(x) = --------------------- = 3(x - 2) = 3x - 6
                 (x + 2)

So, the graph is a right line that intercepts the y-axis at - 6 and the x-axis at x = 2, excluding x = -2 as the function is not defined for x = -2. That is the second graph of the second picture.

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Find the solution to the system of equations-2 + 2y = -4 3x + 3y = -18

AfilCa [17]

Answer:

$\boxed{x = -2~~and~~ y = -4}$

Step-by-step explanation:

-2x + 2y = -4 (times 3) -6x + 6y = -12

3x + 3y = -18 (times 2) 6x + 6y = -36

-6x + 6y = -12

<u>6x + 6y = -36 </u>+

12y = -48

$y = \frac{-48}{12}$

y = -4

Substitute the value of y to the one of equations

-2x + 2(-4) = -4

-2x - 8 = -4

-2x = -4 + 8

-2x = 4

$x = \frac{4}{-2}$

x = -2

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3 years ago

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What is x in this equation

Gennadij [26K]

it would be 81, all interior angles should equal 360 and that goes for any closed shape.

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4 years ago

Select the graph of the solution. Click until the correct graph appears.<br><br> |x| + 3 > 7

JulijaS [17]

Answer:

Graph 3.

Step-by-step explanation:

|x| + 3 > 7

|x| > 4

This means x > 4 or x < -4,

which is the third graph.

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3 years ago

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Rewrite the given parametric equations in Cartesian from x(t) = 2t– 1 y(t) = 5t

docker41 [41]

Hi there!

$\large\boxed{(-\frac{5}{3},-\frac{5}{3})}$

We can begin by setting the two expressions equal to solve for t:

2t - 1 = 5t

Subtract 2t from both sides:

-1 = 3t

Divide both sides by 3:

t = -1/3

Plug in this value of t into x(t) and y(t):

x(-1/3) = 2(-1/3) - 1 = -2/3 - 1 = -5/3

y(-1/3) = 5(-1/3) = -5/3

Thus:

(-5/3, -5/3) are the corresponding coordinates.

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3 years ago

Q1 A ball is thrown upwards with some initial speed. It goes up to a height of 19.6m and then returns. Find (a) The initial spee

lubasha [3.4K]

Answer:

(a) 19.6 ms⁻¹

(b) 2 s

(d) 4 s

Step-by-step explanation:

<u>Constant Acceleration Equations (SUVAT)</u>

$\boxed{\begin{array}{c}\begin{aligned}v&=u+at\\\\s&=ut+\dfrac{1}{2}at^2\\\\ s&=\left(\dfrac{u+v}{2}\right)t\\\\v^2&=u^2+2as\\\\s&=vt-\dfrac{1}{2}at^2\end{aligned}\end{array}} \quad \boxed{\begin{minipage}{4.6 cm}$s$ = displacement in m\\\\$u$ = initial velocity in ms$^{-1}$\\\\$v$ = final velocity in ms$^{-1}$\\\\$a$ = acceleration in ms$^{-2}$\\\\$t$ = time in s (seconds)\end{minipage}}$

When using SUVAT, assume the object is modeled as a particle and that acceleration is constant.

Acceleration due to gravity = 9.8 ms⁻².

When the ball reaches its maximum height, its velocity will momentarily be zero.

<u>Given values</u> (taking up as positive):

$s=19.6 \quad v=0 \quad a=-9.8$

$\begin{aligned}\textsf{Using} \quad v^2&=u^2+2as\\\\\textsf{Substitute the given values:}\\0^2&=u^2+2(-9.8)(19.6)\\0&=u^2-384.16\\u^2&=384.16\\u&=\sqrt{384.16}\\\implies u&=19.6\; \sf ms^{-1}\end{aligned}$

Therefore, the initial speed is 19.6 ms⁻¹.

Using the same values as for part (a):

$\begin{aligned}\textsf{Using} \quad s&=vt-\dfrac{1}{2}at^2\\\\\textsf{Substitute the given values:}\\19.6&=0(t)-\dfrac{1}{2}(-9.8)t^2\\19.6&=4.9t^2\\t^2&=\dfrac{19.6}{4.9}\\t^2&=4\\t&=\sqrt{4}\\\implies t&=2\; \sf s\end{aligned}$

Therefore, the time taken to reach the highest point is 2 seconds.

As the ball reaches its maximum height at 2 seconds, one second before this time is 1 s.

<u>Given values</u> (taking up as positive):

$u=19.6 \quad a=-9.8 \quad t=1$

$\begin{aligned}\textsf{Using} \quad v&=u+at\\\\\textsf{Substitute the given values:}\\v&=19.6+(-9.8)(1)\\v&=19.6-9.8\\\implies v&=9.8\; \sf ms^{-1}\end{aligned}$

The velocity of the ball one second before it reaches its maximum height is the <u>same</u> as the velocity one second after.

<u>Proof</u>

When the ball reaches its maximum height, its velocity is zero.

Therefore, the values for the downwards journey (from when it reaches its maximum height):

$u=0 \quad a=9.8 \quad t=1$

(acceleration is now positive as we are taking ↓ as positive).

$\begin{aligned}\textsf{Using} \quad v&=u+at\\\\\textsf{Substitute the given values:}\\v&=0+9.8(1)\\\implies v&=9.8\; \sf ms^{-1}\end{aligned}$

Therefore, the velocity of the ball one second before <u>and</u> one second after it reaches the maximum height is 9.8 ms⁻¹.

From part (a) we know that the time taken to reach the highest point is 2 seconds. Therefore, the time taken by the ball to travel from the highest point to its original position will also be 2 seconds.

Therefore, the total time taken by the ball to return to its original position after it is thrown upwards is 4 seconds.

4 0

1 year ago