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3 years ago

7

HELP ASAP What are some real world examples you can think of that involve using the coordinate plane? Feel free to use the inter

net to come up with ideas.

1 answer:

sattari [20]3 years ago

4 0

Answer:

scanner and photo copying machines use a coordinate plane to photo copy the exact same image as the origional

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PLEASE HELP! IN DESPERATE NEED! Please SHOW YOUR WORK! Thx so much! :)

choli [55]

The 1st one
y=37
x=18.5

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3 years ago

Can someone answer this question please answer it correctly if it’s corect I will mark you brainliest

Slav-nsk [51]

Answer:

walnut park

Step-by-step explanation:

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3 years ago

Read 2 more answers

Define fn : [0,1] --> R by the

sasho [114]

Answer:

The sequence of functions $\{x^{n}\}_{n\in \mathbb{N}}$ converges to the function

$f(x)=\begin{cases}0&0\leq x$ .

Step-by-step explanation:

The limit $\lim_{n\to \infty }c^{n}$ exists and converges to zero whenever $\lvert c \rvert$ . But, if $c=1$ the sequence $\{c^{n}\}$ is constant and all its terms are equal to $1$ , then converges to $1$ . Using this result, consider the sequence of functions $\{f_{n}\}$ defined on the interval $[0,1]$ by $f_{n}(x)=x^{n}$ . Then, for all $0\leq x$ we have that $\lim_{n\to \infty}x^{n}=0$ . Now, if $x=1$ , then $\lim_{n\to \infty }x^{n}=1$ . Therefore, the limit function of the sequence of functions is

$f(x)=\begin{cases}0&0\leq x$ .

To show that the convergence is not uniform consider $0$ . For any $n>1$ choose $x\in (0,1)$ such that $\varepsilon^{1/n}$ . Then

$\varepsilon$

This implies that the convergence is not uniform.

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3 years ago

you catch 50 fish from the lake tag and release them. teh next day you catch 20 fish and find 6 with tags on them what si teh ex

scZoUnD [109]

seee the answer below

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3 years ago

A parabola can be drawn given a focus of (-9, -7) and a directrix of x = 9. Write

Slav-nsk [51]

Check the picture below, so the parabola looks more or less like so, with a vertex at (0 , -7), let's recall the vertex is half-way between the focus point and the directrix.

so this horizontal parabola opens up to the left-hand-side, meaning that the "P" distance is a negative value.

$\textit{horizontal parabola vertex form with focus point distance} \\\\ 4p(x- h)=(y- k)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h+p,k)}\qquad \stackrel{directrix}{x=h-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{"p"~is~negative}{op ens~\supset}\qquad \stackrel{"p"~is~positive}{op ens~\subset} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\begin{cases} h=0\\ k=-7\\ p=-9 \end{cases}\implies 4(-9)(x-0)~~ = ~~[y-(-7)]^2 \\\\\\ -36x=(y+7)^2\implies x=-\cfrac{1}{36}(y+7)^2$

4 0

2 years ago