Since we have (5,3), f(5)=3
f'(5) is the slope of the line at x=5
we are given that the line tangent to f(x) at x=5 passes through the point (0,2)
so find the slope of that line
we know it passes through (5,3) and (0,2)
slope between

and

is

slope between (5,3) and (0,2) is [/tex]\frac{3-2}{5-0}=\frac{1}{5}[/tex]
f(5)=3
f'(5)=
I think it would be no solution
4x-2(2x+8)=8
4x-4x+16=8
0x+16=8
16=8
which is wrong
Answer:
option Bis correct answer.
Step-by-step explanation:
given, a line passes through the point (_1,3)and has slope 2.let another point be x2 and y2.
by the slope formula,
slope (m)=

or , 2=y2_3/x2_(_1)
or, equating with slope we get,
2=y2_3
or, y2=5
again, 2=x2+1
therefore, x2=1
so the points are (1,5)........ans....
4,295+___=8,329
Answer:4034
Hope this helps, plz make brainly-est!!!