Given that triangle ABC congruent to triangle DEC and m

vodomira [7]

Answer:

Step-by-step explanation:

a) The cost of a product is given as:

Cost = a + bx

Where a is the fixed cost, b is the variable cost, x is the independent variable.

In this case, the independent variable is the distance(d), therefore:

Cost (C) = a + bd

C = a + bd

But the variable cost is $10/km, therefore:

C = a + 10d

A 2.5 km trip cost $40, therefore:

40 = a + 10(2.5)

40 = a + 25

a = 40 - 25 = 15

The fixed cost is $15. Therefore the equation for cost is:

C = 15 + 10d

b) C = 15 + 10d

For 6.5 km ride:

C = 15 + 10(6.5) = 15 + 65 = 80

It would cost $80 for a 6.5 km drive.

7 0

3 years ago

GREYUIT [131]

Answer:

$\displaystyle y' = \frac{-2}{x \ln (10)[\log (x) - 2]^2}$

General Formulas and Concepts:

Calculus

Differentiation

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Derivative Rule [Basic Power Rule]:

Derivative Rule [Quotient Rule]: $\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Step-by-step explanation:

Step 1: Define

Identify.

$\displaystyle y = \frac{\log (x)}{\log (x) - 2}$

Step 2: Differentiate

[Function] Derivative Rule [Quotient Rule]: $\displaystyle y' = \frac{[\log (x) - 2][\log (x)]' - [\log (x) - 2]'[\log (x)]}{[\log (x) - 2]^2}$
Rewrite [Derivative Rule - Addition/Subtraction]: $\displaystyle y' = \frac{[\log (x) - 2][\log (x)]' - [\log (x)' - 2'][\log (x)]}{[\log (x) - 2]^2}$
Logarithmic Differentiation: $\displaystyle y' = \frac{[\log (x) - 2]\frac{1}{\ln (10)x} - [\frac{1}{\ln (10)x} - 2'][\log (x)]}{[\log (x) - 2]^2}$
Derivative Rule [Basic Power Rule]: $\displaystyle y' = \frac{[\log (x) - 2]\frac{1}{\ln (10)x} - \frac{1}{\ln (10)x}[\log (x)]}{[\log (x) - 2]^2}$
Simplify: $\displaystyle y' = \frac{\frac{\log (x) - 2}{\ln (10)x} - \frac{\log (x)}{\ln (10)x}}{[\log (x) - 2]^2}$
Simplify: $\displaystyle y' = \frac{\frac{-2}{\ln (10)x}}{[\log (x) - 2]^2}$
Rewrite: $\displaystyle y' = \frac{-2}{x \ln (10)[\log (x) - 2]^2}$