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garri49 [273]
3 years ago
15

Which point is on the graph of the equation y = 4x - 3?

Mathematics
2 answers:
Mandarinka [93]3 years ago
7 0

Answer:

D. (2, 5)

Step-by-step explanation:

In order to determine whether an ordered pair is on the graph of a given equation, when we plug the x-coordinate in for x and the y-coordinate in for y, the equation should hold true.

Let's try each of the answer choices:

A. (4, -3)

y = 4x - 3

-3 =? 4 * 4 - 3

-3 =? 16 - 3

-3 =? 13

This is clearly wrong, so eliminate A.

B. (3, 4)

y = 4x - 3

4 =? 4 * 3 - 3

4 =? 12 - 3

4 =? 9

This is clearly wrong, so eliminate B.

C. (1, -1)

y = 4x - 3

-1 =? 4 * 1 - 3

-1 =? 4 - 3

-1 =? 1

This is clearly wrong, so eliminate C.

D. (2, 5)

y = 4x - 3

5 =? 4 * 2 - 3

5 =? 8 - 3

5 =? 5

This is true, so D is the answer.

Our answer is thus D.

<em>~ an aesthetics lover</em>

bixtya [17]3 years ago
4 0

Answer:

The answer is D

(2,5)

Step-by-step explanation:

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Rowena can paint a room in $14$ hours, while Ruby can paint it in $6$ hours. If Rowena paints for $x$ hours and Ruby paints for
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Question

Rowena can paint a room in 14 hours, while Ruby can paint it in 6 hours. If Rowena paints for x hours and Ruby paints for y hours, they will finish half of the painting, while if Rowena paints for y hours and Ruby paints for x hours they will paint the whole room. Find the ordered pair (x,y)

Answer:

(x, y) = (231/40, 21/40)

where x = 231/40

y = 21/40

Step-by-step explanation:

From the question, we are told:

Rowena can paint a room in = 14 hours

Ruby can paint it in = 6 hours.

This means

In one hour.

Rowena can paint = 1/14 of the room

Ruby can paint =1/6 hour of the room

From the question, we are told that :

If Rowena paints for x hours and Ruby paints for y hours, they will finish half of the painting

This is represented mathematically as:

(1/14)(x) + ( 1/6)(y) = 1/2...... Equation 1

Also we were told that:

if Rowena paints for y hours and Ruby paints for x hours they will paint the whole room

(1/14)(y) + (1/6)(x) = 1 .......... Equation 2

Bringing the two equations together, we have:

(1/14)(x) + ( 1/6)(y) = 1/2 ....... Equation 1

(1/14)(y) + (1/6)(x) = 1 ............ Equation 2

We find the Lowest common multiple of the numerator 14 and 6 = 42. Hence, we multiply both equations through by 42

3x + 7y = 21 ....... Equation 3

7x + 3y = 42 ......... Equation 4

To solve for x and y in Equation 3 and 4 we would be using the Elimination method

21x + 49y = 147 .......... Equation 5

21x +9y = 126 ........... Equation 6

21x + 49y = 147 .......... Equation 5

-(21x +9y = 126) ........... Equation 6

40y = 21

y = 21/40

To get the value of x , we would substitute 21/40 for y in Equation 3

3x + 7y = 21 ....... Equation 3

3x + 7(21/40) = 21

3x + 147/40 = 21

3x = 21 - 147/40

3x = 693/40

x = (693/40) ÷ 3

x = 693/40 × 1/3

x = 693/120

x = 231/40

Therefore, (x, y) = (231/40, 21/40)

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