f(x)=40+4x+x^2 (e) What is the slope of the chord connecting the points (5, f (5)) and (5 + ∆x, f (5 + ∆x)), when ∆x = 5? Really confused on how to do this and what it means

19

Step-by-step explanation:

So we have the function:

And we want to find the slope of the chord that connects the points (5,f(5)) and (5+Δx, f(5+Δx)).

So first, find the two points.

1) (5,f(5))

Substitute in 5 for x:

Square and multiply:

Add:

So, our first point is (5,85)

2) (5+Δx, f(5+Δx))

Remove the Δx by substituting it with 5. Thus:

Now, substitute in 10 for x. Thus:

Square and multiply:

Add:

So our second point is (10,180).

Now, just find the slope between (5,85) and (10,180) using the slope formula:

Let (5,85) be x₁ and y₁ and let (10,180) be x₂ and y₂. Substitute:

Subtract:

Divide:

So, the slope that connects the chord is 19.

And we are done :)

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2

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step-by-step explanation: