There are 2 tangent lines that pass through the point

and

Explanation:
Given:

The point-slope form of the equation of a line tells us that the form of the tangent lines must be:
![[1]](https://tex.z-dn.net/?f=%5B1%5D)
For the lines to be tangent to the curve, we must substitute the first derivative of the curve for
:



![[2]](https://tex.z-dn.net/?f=%5B2%5D)
Substitute equation [2] into equation [1]:
![[1.1]](https://tex.z-dn.net/?f=%5B1.1%5D)
Because the line must touch the curve, we may substitute 

Solve for x:




± 
±
<em> </em>

There are 2 tangent lines.

and

Answer:
Step-by-step explanation:
the first step in solving this equation is adding 4 on both sides
10d = 95
d = 95/10 = 9.5
Answer:
3.3m+1.4
Step-by-step explanation:
Multiply 5 and 0.9m
Multiply 5 and 0.5
5 times 0.9m is 4.5m
5 times 0.5 is 2.5
Multiply -3 and 0.4m
Multiply -3 and 0.7
-3 times 0.4m is -1.2m
-3 times -.7 is -2.1
4.5m+ 2.5 -1.2m-2.1
Which is 3.3m+1.4
Hope this helps!