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3 years ago

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Steve provides lawn care services in his neighborhood. For each lawn he charges a flat fee of 6 dollars for clean up and 10 doll

ars per hour. Write an equation to represent the relationship the total charge, c, and the number of hours he works, h.

1 answer:

Taya2010 [7]3 years ago

3 0

Answer:

$c(h) = 6 + 10h$

Step-by-step explanation:

We are given the following in the question:

Flat fee = 6 dollars

Charges per hour = 10 dollars per hour

Let Steve work for h hours.

Then, we can write the total charge in the following manner:

Total charge =

$=\text{Flat fee} + (\text{Number of hours}\times \text{Charges per hour})$

Putting values, we get,

$c(h) = 6 + 10h$

is the required equation for total charges if Steve worked for h hours.

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Darlene's dog weighed 5 times as much as Leah's Dog.Together,the dogs weighed 84 pounds. How much did each dog weigh?Write an eq

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Let d = Darlene's dog's weight
let l = Leah's dog's weight
d = 5l
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so, 5l + l = 84, which means 6l= 84
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Please help me answer this question

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The value of dy/dx for the functions are

(i) $\frac{dy}{dx} = 4x^{2}sin2x.cos2x+ 2x. sin^{2}2x$

(ii) $\frac{dy}{dx} =\frac{- y(1+3x^{2})}{2x(1+x^{2}) }$

<h3>Differentiation</h3>

From the question, we are to determine dy/dx for the given functions

(i) $x^{2} sin^{2}2x$

Let $u = x^{2}$

and

$v = sin^{2} 2x$

Also,

Let $w= sin2x$

∴ $v = w^{2}$

First, we will determine $\frac{dv}{dx}$

Using the Chain rule
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$v = w^{2}$

∴ $\frac{dv}{dw} =2w$

Also,

$w= sin2x$

∴ $\frac{dw}{dx} =2cos2x$

Thus,

$\frac{dv}{dx} = 2w \times 2cos2x$

$\frac{dv}{dx} = 2sin2x \times 2cos2x$

$\frac{dv}{dx} = 4sin2x . cos2x$

Now, using the product rule

$\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$

From above

$u = x^{2}$

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Now,

$\frac{dy}{dx} = x^{2} (4sin2x.cos2x)+ sin^{2}2x (2x)$

$\frac{dy}{dx} = 4x^{2}sin2x.cos2x+ 2x. sin^{2}2x$

(ii) $xy^{2}+y^{2}x^{3} +2=0$

Then,

$x.2y\frac{dy}{dx}+ y^{2}(1)+y^{2}.3x^{2} + x^{3}.2y\frac{dy}{dx} +0=0$

$2xy\frac{dy}{dx}+ y^{2}+3x^{2}y^{2} + 2x^{3}y\frac{dy}{dx} =0$

$2xy\frac{dy}{dx}+2x^{3}y\frac{dy}{dx} =- y^{2}-3x^{2}y^{2}$

$\frac{dy}{dx} (2xy+2x^{3}y)=- y^{2}(1+3x^{2})$

$\frac{dy}{dx} =\frac{- y^{2}(1+3x^{2})}{2xy+2x^{3}y}$

$\frac{dy}{dx} =\frac{- y^{2}(1+3x^{2})}{2xy(1+x^{2}) }$

$\frac{dy}{dx} =\frac{- y(1+3x^{2})}{2x(1+x^{2}) }$

Hence, the value of dy/dx for the functions are

(i) $\frac{dy}{dx} = 4x^{2}sin2x.cos2x+ 2x. sin^{2}2x$

(ii) $\frac{dy}{dx} =\frac{- y(1+3x^{2})}{2x(1+x^{2}) }$

Learn more on Differentiation here: brainly.com/question/24024883

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Choose the definition for the function

scZoUnD [109]

Answer:

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Step-by-step explanation:

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FromTheMoon [43]

Answer:

Step-by-step explanation:

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If y = 0, then f(x) = 0 = 1 + .5x

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