we have actually to uncover where (-1,-4) autumn in those points and yes it´s belongs v the very first set so yes the prize is A

an initial one

Step-by-step explanation:

Due to slope type which is y=mx + b.

You are watching: To which graph does the point (−1, 4) belong?

-1 = x and y = 4.

It walk not show the one due to the fact that it would certainly be equal to X, but it does show the negative

1st Graph is correct.

Step-by-step explanation:

Given: suggest ( -1 , -4 )

To find : Graph in i beg your pardon given point belong.

We find by placing given allude in each graph.

Graph 1).

y ≤ -x + 4

LHS = y = -4

RHS = -x + 4 = -(-1) + 4 = 1 + 4 = 5

LHS ≤ RHS.

So, Given point belong come this graph.

Graph 2).

y ≤ -x - 6

LHS = y = -4

RHS = -x - 6 = -(-1) - 6 = 1 - 6 = -5

LHS ≥ RHS.

So, Given allude does not belong to this graph.

Graph 3).

y ≤ 2x - 3

LHS = y = -4

RHS = 2x - 3 = 2(-1) - 3 = -2 - 3 = -5

LHS ≥ RHS.

So, Given allude does no belong come this graph.

Graph 4).

y ≤ 5x - 1

LHS = y = -4

RHS = 5x - 1 = 5(-1) - 1 = -5 - 1 = -6

LHS ≥ RHS.

So, Given suggest does no belong come this graph.

Therefore, 1st Graph is correct.

Option A is correct.

Step-by-step explanation:

Given coordinates of the suggest ( -1 , 4 )

Option A

y ≤ -x + 4

LHS = y = 4

RHS = -x + 4 = -(-1) + 4 = 5

LHS ≤ RHS

So, Given suggest belong to this graph.

Option B

y ≤ -x - 5

LHS = y = 4

RHS = -x - 5 = -(-1) - 5 = -4

LHS ≥ RHS

So, Given suggest does not belong to this graph.

Option C

y ≤ 2x - 3

LHS = y = 4

RHS = 2x - 3 = 2(-1) - 3 = -5

LHS ≥ RHS

So, Given suggest does not belong come this graph.

Option D

y ≤ 5x + 1

LHS = y = 4

RHS = 5x + 1 = 5(-1) + 1 = -4

LHS ≥ RHS

So, Given suggest does not belong to this graph.

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Therefore, option A is correct.

The answer is y ≤ -x + 4 ⇒ the first answer

Step-by-step explanation:

∵ The allude is (-1 , -4)

∵ y ≤ -x + 4 ⇒ -1 ≤ -(-1) + 4

∴ -1 ≤ 5 ⇒ right inequality

If we try the others

y ≤ -x - 6 ⇒ -1 ≤ -(-1) - 6 ⇒ -1 ≤ -5 no true

y ≤ 2x - 3 ⇒ -1 ≤ 2(-1) - 3 ⇒ -1 ≤ -5 no true

y ≤ 5x - 1 ⇒ -1 ≤ 5(-1) - 1 ⇒ -1 ≤ -6 not true

∴ The prize is an initial one

By replacing the allude (-1, 4) right into the very first inequality y 4 4 4lksocossiodks8855  Email

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