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Soloha48 [4]
4 years ago
13

which ordered pair could be added to the relation below to ensure it continues to be a function? (-7,9) (4,-1) (0,5) (-2,-2)

Mathematics
1 answer:
marin [14]4 years ago
4 0

Answer:

[see below]

Step-by-step explanation:

A function's inputs do not repeat. This means that any point with the x-value not repeated with the other points can be added to ensure that it continues as a function.

In this scenario:

{x| x ≠ -7, 4, 0, -2}

A point that does not have the x-value of -7, 0, 4, and -2 could be added to the relation to ensure it continues to be a function.

Hope this helps.

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Answer:

2/3

Step-by-step explanation:

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4 years ago
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Parallelogram ABCD is translated (x + 3, y − 2) and then rotated 90° about the origin in the clockwise direction. Complete the t
Darya [45]

Answer: A″ (−1, 2), B″ (1, 1), C″ (1, −1), D″ (−1, −2)

Step-by-step explanation:

  • there are basically 2 parts in the question.
  • first one is the translation(easier one)

after translation, the corresponding points will be:

A (− 5, 1)   --> a(-5+3,1-2) = (-2,-1)

B (−4, 3)  --> b(-4+3,3-2) = (-1,1)

C (−2, 3)  --> c(-2+3,3-2) = (1,1)

D (−1, 1)    --> d(-1+3,1-2)  = (2,-1)

  • now the second part:

let (x,y) be a point and if it is rotated by "\alpha"

then, the new coordinates (x',y') will be :

x'= xcos\alpha +ysin\alpha

y'=-xsin\alpha +ycos\alpha

here, \alpha = 90 degrees

so,

a(-2,-1)  --> A"(-2cos[90]-1sin[90], -[-2]sin[90]+[-1]cos[90]) =(0-1, 2+0)=(-1,2)

(since, cos[90]=0, sin[90]=1)

b(-1,1)    -->B"(-1cos[90]+1sin[90], -[-1]sin[90]+1cos[90])=(1,1)

c(1,1)     --> C"(1cos[90]+1sin[90],-[1]sin[90]+1cos[90]) = (1,-1)

d(2,-1)   --> D"(2cos[90]+[-1]sin[90],-2sin[90]+[-1]cos[90]) =(-1,-2)

these are the coordinates after both the transformations.

7 0
4 years ago
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10) Megan's started a service club at her high school. After all of the members signed up
bogdanovich [222]

Answer:

Your answer is 55

Step-by-step explanation:

11% of 500 is 55

hope it helps :)

pls mark brainliest :P

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3 years ago
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marishachu [46]
-40 4 10 20 I think this is the answer I’m not sure
5 0
3 years ago
G(g(x)) g=(x+14)/(x+2)
GalinKa [24]

\bf g(x)=\cfrac{x+14}{x+2}~\hspace{8em}g(~~g(x)~~)=\cfrac{g(x)+14}{g(x)+2}\\\\\\(~~g(x)~~)=\cfrac{~~\frac{x+14}{x+2}+14~~}{\frac{x+14}{x+2}+2}\implies (~~g(x)~~)=\cfrac{~~\frac{x+14~~+14x+28}{x+2}~~}{\frac{x+14~~+2x+4}{x+2}}\\\\\\(~~g(x)~~)=\cfrac{~~\frac{15x+42}{x+2}~~}{\frac{3x+18}{x+2}}\implies (~~g(x)~~)=\cfrac{15x+42}{\underline{x+2}}\cdot \cfrac{\underline{x+2}}{3x+18}\\\\\\(~~g(x)~~)=\cfrac{15x+42}{3x+18}\implies (~~g(x)~~)=\cfrac{3(5x+14)}{3(x+6)}\implies (~~g(x)~~)=\cfrac{5x+14}{x+6}

6 0
3 years ago
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