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3 years ago

What is the slope of the line passing through the point A and B, as shown on the graph below

Mathematics

1 answer:

IRISSAK [1]3 years ago

8 0

Answer:

We do not see the image

Step-by-step explanation:

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Slope (-3,0 2,0 3,4 )

PtichkaEL [24]

Answer:

undefined

Step-by-step explanation:

so you can just pick 2 points. use -3,0, and 2,0. so to get from -3 to 2(becusse slope is change of y over change of x) you add 5. to get from 0 to 0, you would add zero. so, 5/0=undefined. I'm not quite sure why thered 3,4. in that case it may just not be a function

3 0

4 years ago

Round 5984 to the nearest hundred.a) 5000 b) 5900 c) 5980 d) 5990 e) 6000

SpyIntel [72]

The answer would be E. because if the determining number (The number directly after the place being rounded to. The 8 in this case) is greater than five, you round up and drop the numbers after the place being rounded to (The hundredth's place), if it's less, the number in the place being rounded to stays the same, but you still drop the numbers after it.

4 0

3 years ago

Read 2 more answers

What does a slope of 1/4 look like

Reil [10]

Slope is also known as rise over run -
$\frac{rise}{run}$
Here, we have the slope $\frac{1}{4}$ . This means that the rise is $1$ and the run is $4$ . Since both values are positive, we can say that we go up one and we go right four. When we plot this on a graph (the graph is attached), we get the same exact formula we found. **In the graph, I plotted the line $y= \frac{1}{4} x$ **. Hope this helped!

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.

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