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GuDViN [60]
2 years ago
11

Select the correct answer.

Mathematics
1 answer:
Tanya [424]2 years ago
8 0

Answer:

The function f has an inverse because it is one-to-one.

Step-by-step explanation:

The condition for a function to have an inverse is that every x coordinate maps to a single y - coordinate. As illustrated by the diagram attached in the second picture, every x value maps to one y value therefore satisfying the condition making the function have an inverse.

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Uestion
Stella [2.4K]

Check the picture below, so the park looks more or less like so, with the paths in red, so let's find those midpoints.

~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ J(\stackrel{x_1}{-3}~,~\stackrel{y_1}{1})\qquad K(\stackrel{x_2}{1}~,~\stackrel{y_2}{3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 1 -3}{2}~~~ ,~~~ \cfrac{ 3 +1}{2} \right) \implies \left(\cfrac{ -2 }{2}~~~ ,~~~ \cfrac{ 4 }{2} \right)\implies JK=(-1~~,~~2) \\\\[-0.35em] ~\dotfill

~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ L(\stackrel{x_1}{5}~,~\stackrel{y_1}{-1})\qquad M(\stackrel{x_2}{-1}~,~\stackrel{y_2}{-3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ -1 +5}{2}~~~ ,~~~ \cfrac{ -3 -1}{2} \right) \implies \left(\cfrac{ 4 }{2}~~~ ,~~~ \cfrac{ -4 }{2} \right)\implies LM=(2~~,~~-2) \\\\[-0.35em] ~\dotfill

~~~~~~~~~~~~\textit{distance between 2 points} \\\\ JK(\stackrel{x_1}{-1}~,~\stackrel{y_1}{2})\qquad LM(\stackrel{x_2}{2}~,~\stackrel{y_2}{-2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ JKLM=\sqrt{(~~2 - (-1)~~)^2 + (~~-2 - 2~~)^2} \\\\\\ JKLM=\sqrt{(2 +1)^2 + (-2 - 2)^2} \implies JKLM=\sqrt{( 3 )^2 + ( -4 )^2} \\\\\\ JKLM=\sqrt{ 9 + 16 } \implies JKLM=\sqrt{ 25 }\implies \boxed{JKLM=5}

now, let's check the other path, JM and KL

~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ J(\stackrel{x_1}{-3}~,~\stackrel{y_1}{1})\qquad M(\stackrel{x_2}{-1}~,~\stackrel{y_2}{-3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ -1 -3}{2}~~~ ,~~~ \cfrac{ -3 +1}{2} \right) \implies \left(\cfrac{ -4 }{2}~~~ ,~~~ \cfrac{ -2 }{2} \right)\implies JM=(-2~~,~~-1) \\\\[-0.35em] ~\dotfill

~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ K(\stackrel{x_1}{1}~,~\stackrel{y_1}{3})\qquad L(\stackrel{x_2}{5}~,~\stackrel{y_2}{-1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 5 +1}{2}~~~ ,~~~ \cfrac{ -1 +3}{2} \right) \implies \left(\cfrac{ 6 }{2}~~~ ,~~~ \cfrac{ 2 }{2} \right)\implies KL=(3~~,~~1) \\\\[-0.35em] ~\dotfill

~~~~~~~~~~~~\textit{distance between 2 points} \\\\ JM(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-1})\qquad KL(\stackrel{x_2}{3}~,~\stackrel{y_2}{1})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ JMKL=\sqrt{(~~3 - (-2)~~)^2 + (~~1 - (-1)~~)^2} \\\\\\ JMKL=\sqrt{(3 +2)^2 + (1 +1)^2} \implies JMKL=\sqrt{( 5 )^2 + ( 2 )^2} \\\\\\ JMKL=\sqrt{ 25 + 4 } \implies \boxed{JMKL=\sqrt{ 29 }}

so the red path will be  5~~ + ~~\sqrt{29} ~~ \approx ~~ \blacksquare~~ 10 ~~\blacksquare

3 0
2 years ago
Solve the system of equations. \begin{aligned} & -10x+3y = 5 \\\\ & x=y-4 \end{aligned} ​ −10x+3y=5 x=y−4 ​
Ksenya-84 [330]

Answer:

x = 1

y = 5

Step-by-step explanation:

You can use either substitution or elimination for this problem. I will use substitution.

Step 1: Replace x in the 1st equation

-10(y-4) + 3y = 5

-10y + 40 + 3y = 5

-7y = -35

y = 5

Step 2: Plug <em>y </em>into an original equation to find <em>x</em>

x = 5 - 4

x = 1

And you have your final answers!

7 0
3 years ago
Need help!!!!ASAP
raketka [301]
I hope this helps you

3 0
3 years ago
Which is an equivalent ratio to 6/7?​
anastassius [24]

Answer: 12 : 14  18 : 21  24 : 28  30 : 35

Step-by-step explanation:

hope these is the right answer to your question

6 0
3 years ago
NEED YOUR HELP!! I WILL UPVOTE!
vodomira [7]
12x=-48 (divide 12 from both sides)
x=-4

10y-13y=30
-3y=30 (divide -3 from both sides)
-10=y

10x+14x=48
24x=48 (divide 24 from both sides)
x=2


5 0
3 years ago
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