Answer:
Step-by-step explanation:
His English class can be located on coordinates-3,-2
The next answer is (-3,-7)
Answer:
C srry if wrong
Step-by-step explanation:
Answer:
I am pretty sure Its C-120
Step-by-step explanation:
Hope This Helps
d=
Let's simplify step-by-step.
p+6pq−2pq−129
=p+6pq+−2pq+−129
Combine Like Terms:
=p+6pq+−2pq+−129
=(6pq+−2pq)+(p)+(−129)
=4pq+p+−129
Answer:
=4pq+p−129
g=
Let's simplify step-by-step.
2x−3y+b(3y−2x)+2.718282−(2x−3y)
Distribute the Negative Sign:
=2x−3y+b(3y−2x)+2.718282+−1(2x−3y)
=2x+−3y+b(3y−2x)+2.718282+−1(2x)+−1(−3y)
=2x+−3y+b(3y−2x)+2.718282+−2x+3y
Distribute:
=2x+−3y+(b)(3y)+(b)(−2x)+2.718282+−2x+3y
=2x+−3y+3by+−2bx+2.718282+−2x+3y
Combine Like Terms:
=2x+−3y+3by+−2bx+2.718282+−2x+3y
=(−2bx)+(3by)+(2x+−2x)+(−3y+3y)+(2.718282)
=−2bx+3by+2.718282
Answer:
=−2bx+3by+2.718282
Okay I think there has been a transcription issue here because it appears to me there are two answers. However I can spot where some brackets might be missing, bear with me on that.
A direct variation, a phrase I haven't heard before, sounds a lot like a direct proportion, something I am familiar with. A direct proportion satisfies two criteria:
The gradient of the function is constant s the independent variable (x) varies
The graph passes through the origin. That is to say when x = 0, y = 0.
Looking at these graphs, two can immediately be ruled out. Clearly A and D pass through the origin, and the gradient is constant because they are linear functions, so they are direct variations.
This leaves B and C. The graph of 1/x does not have a constant gradient, so any stretch of this graph (to y = k/x for some constant k) will similarly not be direct variation. Indeed there is a special name for this function, inverse proportion/variation. It appears both B and C are inverse proportion, however if I interpret B as y = (2/5)x instead, it is actually linear.
This leaves C as the odd one out.
I hope this helps you :)
