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Alexxx [7]
2 years ago
15

Does (-1, 2), (1, 1),( 1 -1), (2,1), (4,2) represent y as a function

Mathematics
1 answer:
Paha777 [63]2 years ago
8 0

No. A function maps one input to exactly one output. The given relation maps 1 to both 1 and -1, as indicated by the pairs (1, 1) and (1, -1).

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The numerator and denominator of a fraction are in the ratio of 3 to 5. If the numerator and denominator are both increased by 2
butalik [34]

Answer:

The correct answer is A) 5n = 3d and 3n + 6 = 2d + 4

Step-by-step explanation:

hope this is right...

6 0
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5 eights greater than 2 thirds
beks73 [17]
No 5/8 is not greater then 2/3, 5/8 is just 1/24 smaller then 2/3.
4 0
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I need help on this asap brainliest to first person
dimulka [17.4K]

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i think 30° gonna stay the same

Step-by-step explanation:

6 0
2 years ago
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Select the correct answer. Find the solution(s) for x in the equation below.<br> x2-9x+20=0
trasher [3.6K]

Step-by-step explanation:

Given

x² - 9x + 20 = 0

x² - ( 4 + 5) x + 20 = 0

x² - 4x - 5x + 20 = 0

x ( x - 4 ) - 5 (x - 4 ) = 0

(x - 4 ) ( x - 5 ) = 0

Either

x - 4 =0

x = 4

or

x - 5 =0

x = 5

x = 4 and 5

Hope it will help :)

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8 0
2 years ago
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The equation of a circle is x^2+y^2-2x +6y-6=0. Find the center and radius of the circle by completing the square.you must inclu
zavuch27 [327]

Answer:

The center of the circle is at;

(1,-3)

The radius of the circle is;

r=4

Explanation:

Given the equation of circle;

x^2+y^2-2x+6y-6=0

we want to re-write it in the form;

(x-h)^2+(y-k)^2=r^2

where;

\begin{gathered} (h,k)\text{ is the center of the circle} \\ r\text{ is the radius} \end{gathered}

Applying Completing the square method;

\begin{gathered} x^2+y^2-2x+6y-6=0 \\ x^2-2x+1+y^2+6y+9=6+1+9 \\ (x-1)^2+(y+3)^2=16 \\ (x-1)^2+(y+3)^2=4^2 \end{gathered}

comparing the derived equation to the general form we have;

\begin{gathered} h=1 \\ k=-3 \\ r=4 \end{gathered}

Therefore;

The center of the circle is at;

(1,-3)

The radius of the circle is;

r=4

6 0
9 months ago
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