2 3 x + 1 2 y = 2 3 8 x + 5 6 y = − 11 2

PLEASE HELP :’( last one

valina [46]

Answer:

A.3

Step-by-step explanation:

5 0

2 years ago

https://browardcountyschools-my.sharepoint.com/:w:/g/personal/0617013669_my_browardschools_com/EQC_jBr4uQlFlCStDW2aK0ABF8ld3ETXM

anzhelika [568]

Answer:

ty i went there Dont Have A Miscroft account I Have One but My School Locked it

Step-by-step explanation:

6 0

2 years ago

A line parallel to the y-axis could contain which point? Choose all that apply.

madreJ [45]

The answer is b and c
because they are both parallel to the y-axis

5 0

2 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:

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

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

2 years ago

What is the range and domain of the function p(t)=-8t+100

Aleks [24]

Answer:

Real numbers for both

Step-by-step explanation:

The domain of a function is the set of values that the unknown t can adopt. For this function, t can be any real number as there are no restrictions for the t. Ir can be any positive number, 0, negative numbers, fractions, irrational numbers, whatever number you like.

The range of a function is the values that p(t) adopt when we replace the t value with any number. Here, again, it range is all real numbers. If you want p(t) to be positive it is possible, negative is possible, 0 is possible, and so on. If you like, you can verify it by replacing the numbers you like.

Something to know is that linear polynomial functions ALWAYS have their domains and ranges in real numbers.

5 0

2 years ago