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. Clearly A and D pass through the origin, and the gradient is constant because they are linear functions, so they are direct variations. 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. I believe the answer is C. Hope I helped!
Answer:
A function is a type of <em><u>vertical line </u></em><em><u>test.</u></em>
Answer:
Step 2:
For Completing the square,
Add half coefficient of x square on both the side we get
Step-by-step explanation:
Solve:
Solution:
Step 1:
Dividing both the side by two we get
Step 2:
For Completing the square,
Add half coefficient of x square on both the side we get
Step 3:
We know 
<span>in standard form the equation has to be changed from slope-intercept form to standard. bt first you have to find the slope and then change the equation to standard form.
the slope intercet form is y=mx+b using y-k= m(x-h)
and after, change the equation to ax+by+c</span>
Answer:
5
Step-by-step explanation:
