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ad-work [718]
3 years ago
12

Which graph represents the function r(x) = |x – 2| - 1​

Mathematics
1 answer:
maria [59]3 years ago
5 0

Answer:you diddnt attach a graph but it woul have a stating pooint of positive 3 and a slove of up one over 1  so 1/1 +3

Step-by-step explanation:

You might be interested in
Write the equation 9y = 12x + 0.2 in standard form. Identify A, B, and C.
uysha [10]

Answer:

A = -12, B = 9 and C = .2

Step-by-step explanation:

So it just looks like we need to use algebra to move everything around to get it in the form Ax + By = C

So, let's start with 9y = 12x + 0.2.  From standard form we have all terms wth variables on one side, so let's do that.  the right sie has both a variable term and non variable term, so let's get rid of the variable term.  The variable term is 12x, so to get rid of it you subtract 12x from both sides.

9y = 12x + 0.2

9y - 12x = 0.2

To make it exactly in standard form just rarrnge.

-12x + 9y = 0.2

So now we have A = -12, B = 9 and C = .2

5 0
3 years ago
Anyone know how to do this?
liraira [26]

Answer: It adding 6.

Step-by-step explanation: -18 plus 6 equals -12. -12 plus 6 = -6 and so on.

4 0
2 years ago
Read 2 more answers
HELP WHAT IS THE SLOPE OF THIS GRAPH!!! DUE TODAY ALMOST LATE!!!!!
Nady [450]

Answer:

wait what?

how is something meant to be due late, like b r u h you can't make a task due late. LOL :)

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
Can someone check if my answers are right? I got 6,-18,54,-162,486,1458​
Lyrx [107]

Answer:

yes, you have it right except you need the 1458 negative

Step-by-step explanation:

Nice job getting them right!

8 0
2 years ago
AABC has vertices at A(1, -9), B(8,0), and C(9,-8).
Rainbow [258]

Check the picture below.

~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{-9})\qquad B(\stackrel{x_2}{8}~,~\stackrel{y_2}{0}) ~\hfill AB=\sqrt{[ 8- 1]^2 + [ 0- (-9)]^2} \\\\\\ AB=\sqrt{7^2+(0+9)^2}\implies AB=\sqrt{7^2+9^2}\implies \boxed{AB=\sqrt{130}} \\\\[-0.35em] ~\dotfill

B(\stackrel{x_1}{8}~,~\stackrel{y_1}{0})\qquad C(\stackrel{x_2}{9}~,~\stackrel{y_2}{-8}) ~\hfill BC=\sqrt{[ 9- 8]^2 + [ -8- 0]^2} \\\\\\ BC=\sqrt{1^2+(-8)^2}\implies \boxed{BC=\sqrt{65}}

now, we could check for the CA distance, however, we already know that AB ≠ BC, so there's no need.

6 0
2 years ago
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