Write an equation of a line which has a slope of -2 and goes through the point (-4, 11).

y -11 = -2(x+4)

or

y = -2x+3

Step-by-step explanation:

We can use the point slope form of a line

y - y1 = m(x-x1)  where m is the slope and (x1 ,y1) is a point on the line

y - 11 = -2(x- -4)

y -11 = -2(x+4)

We can rewrite it in slope intercept form

y-11 = -2x-8

Add 11 to each side

y = -2x+3

the equation would be y=-2x+3

Step-by-step explanation:

you are given the slope and a point on this line, and the form for an equation of a line is y=mx+b, where m is the slope and b is the y-intercept. so since we know the slope already:

y=-2x+b

and we are given a point, (x, y), so plug in those numbers as x and y.

11=-2(-4)+b

now solve for b:

11=8+b

3=b

now plug in the slope, m, and y-intercept, b, back into the original equation:

so, y=-2x+3

4th option is correct xy=-10

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δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)