A and B lie on the line, yes, but what specifically are you supposed to do? Looks like your problem statement was cut off before you'd finished typing it in.
You say your line passes thru (-2,5) and has a slope of 2/3? Then, using the point-slope formula,
y-5 = (2/3)(x+2) This is the general equation for your line.
Now let's play around with B(-2,y). Suppose we subst. the x-coordinate of B, which is -2, into the equation y-5 = (2/3)(x+2); we get y-5 = (2/3)(-2+2) = 0. This tells us that y must be 5. But we already knew that!!
So, please review the original problems with its instructions and this discussion and tell me what you need to know from this point on.
The correct answer is 50:1 I got that one right!
Answer:
See explanation
Step-by-step explanation:
First plot points A, B, C and D on the coordinate plane and connect them to get quadrilateral ABCD. Then plot the vertices E, F, G, H of the quadrilateral EFGH in each case.
For each of these cases there are attached diagrams
1 case - reflection acraoss the x-axis
2 case - translation 5 units down
3 case - reflection across the y-axis
4 case - translation 7 units right
Ln(56) = 4x-9
ln(56) + 9 = 4x
(ln(56) + 9)/4 = x
Answer:
{- 2, - 4, - 6, - 8, - 10 }
Step-by-step explanation:
Given
f(x) = 2x - 6 with domain { - 2, - 1, 0, 1, 2 }
To obtain the range substitute the values of x from the domain into f(x)
f(- 2) = 2(- 2) - 6 = - 4 - 6 = - 10
f(- 1) = 2(- 1) - 6 = - 2 - 6 = - 8
f(0) = 2(0) - 6 = 0 - 6 = - 6
f(1) = 2(1) - 6 = 2 - 6 = - 4
f(2) = 2(2) - 6 = 4 - 6 = - 2
Range is { - 2, - 4, - 6, - 8, - 10 }
