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Explanation:
10. Jack and Jill were working on their homework together but ended up with two different
answers to the same problem. The problem asked for a new line, parallel to y=2x+3, and
containing the point (-2,-2). Who got the problem correct? Explain your reasoning?
Jack:
y-2 = 2(x - 2)
y-2 = 2x - 4
y = 2x - 2
Jill:
y + 2 = 2(x + 2)
y + 2 = 2x + 4
y = 2x + 2
Answer:
Explanation:

Explanation 10 Jack and Jill were working on their homework together but ended up with two different answers to the same problem The problem asked for a new lin class=

Respuesta :

Step-by-step explanation:

Hey there!

Firstly we need to find an equation of the line passing through point (-2,-2).

Now,

The equation of a st.line passing through point (-2,-2) is,

[tex](y - y1) = m1(x - x1)[/tex]

Now, putting the values,

[tex](y + 2) = m1(x + 2)[/tex]

It is first equation.

Now, another equation is,

y= 2x+3

Comparing, the equation with y = mx+c.

slope (m2)= 2

As per the condition of parallel lines,

m1=m2= 2

Putting, the values of m1 in equation (i)

(y+2)= 2(x+2)

y= 2x+2....is the equation.

Therefore, Jill's equation is correct answer.

Hope it helps...

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