we can always find the x-intercept by simply settting y = 0, and solving for "x".
and we can always find the y-intercept by simply setting x = 0 and solving for "y".
![\bf x-4y=-16\implies \stackrel{x=0}{0-4y=-16}\implies y=\cfrac{-16}{-4}\implies y=4 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill (0,4)~\hfill](https://tex.z-dn.net/?f=%20%5Cbf%20x-4y%3D-16%5Cimplies%20%5Cstackrel%7Bx%3D0%7D%7B0-4y%3D-16%7D%5Cimplies%20y%3D%5Ccfrac%7B-16%7D%7B-4%7D%5Cimplies%20y%3D4%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A~%5Chfill%20%280%2C4%29~%5Chfill%20)
Answer:-3 3/5 <1.3
Step-by-step explanation:
Answer:
- f¯¹(x) =9x+18, for the pictorial(image) question
- what is the send question is it asking derivative or inverse
- If it is inverse f¯¹(x)=-x-2 as it is
- Or if it is derivative f'(x)=-1
Step-by-step explanation:
<h2>
For the image question f(x)=1/9x-2,f¯¹(x)=?</h2>
- f(x)=1/9x-2............given
- y=1/9x-2................swapping f(x) by y to Easily write
- x=1/9y-2................interchanging x and y
- 1/9y-2=x................changeling side of equation
- 9(1/9y-2)=(x)9.......multiplying both sides by 9 to override the fraction on the right side
- y-18=9x
- y=9x-18.................Return to where it were
- <em><u>f¯¹(x)=9x+18</u></em>..........swap back f¯¹(x) in the y
<h2>
For the question f(x)=-x-2,f¯¹(x)=?</h2>
- following the ☝️ arrangement
- y=-x-2
- x=-y-2
- -y-2=x
- -y=x+2
- y=-x-2
- f¯¹(x)=-x-2
C the answer hehehehehehehwhw
Mode are repeating numbers in a sequence therefore 23 is your mode
