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3 years ago

Calculus help needed!

Mathematics

1 answer:

Westkost [7]3 years ago

5 0

Answer:

Not continous at x = 6

Step-by-step explanation:

When x = 6

G(6) = 1/(6 - 6) = 1/0

Dividing by 0 creates a discontinuity

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Find the sine of angle S. Reduce the answer to the lowest terms.

NeX [460]

Answer:

$\sin e\text{ S = }\frac{3}{5}$

Explanation:

Here, we want to get the value of sine S

The sine of an angle is the ratio of the opposite side to the hypotenuse side

The hypotenuse side is the side that faces the right angle which is the small box. It is the longest side of the triangle and it measures 10

The opposite is the side that faces the angle which is 6

We have the sine as:

$\frac{6}{10}\text{ = }\frac{3}{5}$

8 0

1 year ago

The equation of the graphed line is 2x – 3y = 12.

Viefleur [7K]

Answer:

Step-by-step explanation:

The x-intercept is the x-coordinate of the point on the line that intersects the x-axis. At the x-axis, all y-coordinates are 0. Let y = 0, and solve for x.

2x – 3y = 12

2x - 3(0) = 12

2x = 12

x = 6

6 0

3 years ago

Read 2 more answers

At the farmers marke, two pounds of peaches cost $4.20. How much will five pounds cost

Paha777 [63]

10.50 should be your answer

6 0

3 years ago

Read 2 more answers

HI PLASE HELP -3x-4<5

Dmitry_Shevchenko [17]

Answer:

-3

Step-by-step explanation:

-3x - 4 < 5

-3x < 9

x < -3

3 0

3 years ago

F(x)=x-1/x^2-x-6 which is the graph of

hoa [83]

Answer:

The graph is attached below.

Step-by-step explanation:

<em>As you have not added the graph, so I will be solving the function for a graph.</em>

Given the function

$f\left(x\right)=x-\frac{1}{x^2}-x-6$

$x-\mathrm{axis\:interception\:points\:of\:}-\frac{1}{x^2}-6:$

$\mathrm{x-intercept\:is\:a\:point\:on\:the\:graph\:where\:}y=0$

$-\frac{1}{x^2}-6=0$

$-1-6x^2=0$

$\mathrm{No\:Solution\:for}\:x\in \mathbb{R}$

$\mathrm{No\:x-axis\:interception\:points}$

$y-\mathrm{axis\:interception\:point\:of\:}-\frac{1}{x^2}-6:$

$y\mathrm{-intercept\:is\:the\:point\:on\:the\:graph\:where\:}x=0$

As we know that the domain of a function is the set of input or argument values for which the function is real and defined.

$\mathrm{Domain\:of\:}\:-\frac{1}{x^2}-6\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x0\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:0\right)\cup \left(0,\:\infty \:\right)\end{bmatrix}$

$\mathrm{Since}\:x=0\:\mathrm{is\:not\:in\:domain}$

$\mathrm{No\:y-axis\:interception\:point}$

$\mathrm{Asymptotes\:of}\:-\frac{1}{x^2}-6:\quad \mathrm{Vertical}:\:x=0,\:\mathrm{Horizontal}:\:y=-6$

$\mathrm{Range\:of\:}-\frac{1}{x^2}-6:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)$

The graph is attached below.

5 0

3 years ago