-$69.68 is your answer if I hope that this is correct.
Answer:
I think I can help, whats your quiz over in algebra?
Answer:
Below
I hope its not too complicated

Step-by-step explanation:




X =(-8+√464)/-8=1-1/2√<span> 29 </span><span>= -1.693</span>
Answer:
A. y + 2= x
Step-by-step explanation:
Which equation is represented by the graph shown in the image?
A. y + 2= x
B. y + 1= x
C. y - 1= x
D. y - 2= x
Please show ALL work! <3
The graph shown has a slope of +1 and a y intercept of -2.
All given answer choices have a slope of +1, so that's not the problem.
We need one that has a y-intercept of -2, or the equation should be
y = x-2, or equivalently y+2 = x
which corresponds to answer choice A.