Answer:
First, solve for two points as an equation instead of an inequality to find the boundary line for the inequality.
For:
x
=
0
y
=
0
+
5
y
=
5
Or
(
0
,
5
)
For:
x
=
−
2
y
=
−
2
+
5
y
=
3
Or
(
−
2
,
3
)
We can now plot the two points on the coordinate plane and draw a line through the points to mark the boundary of the inequality.
The boundary line will be solid because the inequality operator contains an "or equal to" clause.
graph{(x^2+(y-5)^2-0.125)((x+2)^2+(y-3)^2-0.125)(y-x-5)=0 [-20, 20, -10, 10]}
Now, we can shade the left side of the line.
graph{(y-x-5) >= 0 [-20, 20, -10, 10]}
The number of the day is 60 because the LCM is 30
Answer:
Neither, why... for it to be direct or inverse variation, the model has to fit either y=k/x or y=kx, it may not have a y-intercept at all if it is inverse variation and it must have a y-intercept of 0 for it to be direct variation.
Step-by-step explanation:
The statement y=2 is quite specific. Because k is positive, y increases as x increases. So as x increases by 1, y increases by 1.5. Inverse Variation: Because k is positive, y decreases as x increases. yx is a constant number -8. The constant of variation, k , is 23 . Inverse Variation
An inverse variation can be represented by the equation xy=k or y=kx .
That is, y varies inversely as x if there is some nonzero constant k such that, xy=k or y=kx where x≠0,y≠0 .
Suppose y varies inversely as x such that xy=3 or y=3x . That graph of this equation shown.
Hello there!
(-1)^3 * (-1)^2
Use the distributive property
= (-1)(-1)(-1)* (-1)(-1)
= (-1)^5
= -1
I hope this helps!
