f(x)=lnx
y=f(x)
dy/dx= 1/x
tangent at (x,y) has slope 1/x
eqn of tangent is y = mx + c
since the tangent passes through origin, c=0
substitute y = lnx and m= 1/x to above eqn
lnx = 1
x=e
y=lne=1
Zack reads faster so he will finish first
Pam= 126/3 = 42/1
Zack= 180/4 =45/1
Now multiply the unit rates by 5 each
Pam= 210/5
Zack= 225/5
The answer is f(x)=x²+2x when evaluated with -3 gives you the value of 3
Let's check all functions.
1. The function f(x)=x²<span>+2x when evaluated with 3 gives you the value of 3:
Evaluated with x means that</span> x = 3.
f(3) = 3² + 2 * 3 = 9 + 6 = 15
15 ≠ 3, so, this is not correct.
2. f(x)=x²<span>-3x when evaulated with -3 give you the value of 3
Evaluated with -3 means that x = -3.
(f-3) = (-3)</span>² - 3 * (-3) = 9 + 9 = 18
18 ≠ 3, so, this is not correct.
3. f(x)=x²<span>+2x when evaluated with -3 gives you the value of 3
</span> Evaluated with -3 means that x = -3.
f(-3) = (-3)² + 2 * (-3) = 9 - 6 = 3
3 = 3, so this is correct.
4. f(x)=x²-3x when evaluated with -3 gives you the value of 3
Evaluated with 3 means that x = 3.
f(3) = (3)² - 3 * 3 = 9 - 9 = 0
0 ≠ 3, so this is not correct.
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The answer to this would be C because you can multiply everything by 3 and still have a consistent-dependent system.