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Anvisha [2.4K]
3 years ago
9

Which table of ordered pairs represents a linear function? x y −6 26 0 3 6 26 12 101 x y −6 −3 0 3 6 1 12 2 x y −6 −3 0 3 6 9 6

3 x y −6 −3 0 2 6 7 12 12
Mathematics
2 answers:
Afina-wow [57]3 years ago
7 0

In each table, the x-values differ by 6. A linear function will also have a constant difference of y-values. Only the 4th table has that.

(x, y) = (-6, -3), (0, 2), (6, 7), (12, 12)

_____

An increase in x by 6 produces an increase in y by 5. The slope of this linear function is 5/6. In slope-intercept form, the equation is y = (5/6)x + 2.

zubka84 [21]3 years ago
6 0

Answer:

D:  

<u>x   y </u>

<u>−6 −3 </u>

<u>0   2 </u>

<u>6   7 </u>

<u>12   12 </u>

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  arc AC = 63°

Step-by-step explanation:

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18 hours

Step-by-step explanation:

if 10 people spend 15 hours

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Read 2 more answers
For each value of y, determine whether it is a solution to 12 = 2y +8.<br> Is it a solution
grin007 [14]

Answer: 2(0) + 8 does not equal 12, not a solution.

2(2) +8 = 12 yes it is a solution

2(-3) + 8 does not equal 12, not a solution

2(5) + 8 does not equal 12, not a solution.

Step-by-step explanation:

Looks like you need to plug in each y value given and multiplied by 2 and add 8

2(0) + 8 does not equal 12, not a solution.

2(2) +8 = 12 yes it is a solution

2(-3) + 8 does not equal 12, not a solution

2(5) + 8 does not equal 12, not a solution.

3 0
3 years ago
¿Cuál de los siguientes puntos no pertenece a la función cuadrática f(x) = 1-x^2 ?
dusya [7]

Answer:

<h2>A. (0,1)</h2>

Step-by-step explanation:

The question lacks the e=required option. Find the complete question below with options.

Which of the following points does not belong to the quadratic function

f(x) = 1-x²?

a.(0,1) b.(1,0) c.(-1,0)

Let f(x) = 0

The equation becomes 1-x² = 0

Solving 1-x² = 0 for x;

subtract 1 from both sides;

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multiply both sides by minus sign

-(-x²) = -(-1)

x² = 1

take square root of both sides;

√x² = ±√1

x = ±1

x = 1 and x = -1

when x = 1

f(x) = y = 1-1²

y = 1-1

y = 0

when x = -1

f(x) = y = 1-(-1)²

y = 1-1

y = 0

Hence the coordinate of the function f(x) = 1-x² are (±1, 0) i.e (1, 0) and (-1, 0). The point that does not belong to the quadratic function is (0, 1)

3 0
2 years ago
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