Jesse looks at the prices of 5 guitars. The prices are $85,

Finding Derivatives Implicity In Exercise,Find dy/dx implicity.<br> x2e - x + 2y2 - xy = 0

Klio2033 [76]

Answer:

the question is incomplete, the complete question is

"Finding Derivatives Implicity In Exercise,Find dy/dx implicity . $x^{2}e^{-x}+2y^{2}-xy$ "

Answer : $\frac{dy}{dx}=\frac{y-(2-x)xe^{-x}}{(4y-x)}$

Step-by-step explanation:

From the expression $x^{2}e^{-x}+2y^{2}-xy$ " y is define as an implicit function of x, hence we differentiate each term of the equation with respect to x.

we arrive at

$\frac{d}{dx}(x^{2}e^{-x )+\frac{d}{dx} (2y^{2})-\frac{d}{dx}xy=0\\$

for the expression $\frac{d}{dx}(x^{2}e^{-x})$ we differentiate using the product rule, also since y^2 is a function of y which itself is a function of x, we have

$(2xe^{-x}-x^{2}e^{-x})+4y\frac{dy}{dx}-x\frac{dy}{dx} -y=0\\\\(2-x)xe^{-x}+(4y-x)\frac{dy}{dx}-y=0 \\$ .

if we make dy/dx subject of formula we arrive at

$(4y-x)\frac{dy}{dx}=y-(2-x)xe^{-x}\\\frac{dy}{dx}=\frac{y-(2-x)xe^{-x}}{(4y-x)}$

5 0

3 years ago

Melissa Costouras obtains a $3,000 loan for darkroom equipment. She makes six monthly payments of $511.18. Determine the APR.

nasty-shy [4]

Using the simple interest formula, it is found that the APR for the loan is of 4.472%.

<h3>What is the simple interest formula and when it is used?</h3>

Simple interest is used when there is a single compounding per time period.

The amount of money after t years in is modeled by:

$A(t) = A(0)(1 + rt)$

In which:

A(0) is the initial amount.

r is the interest rate, as a decimal.

The parameters for this problem are:

A(t) = 6 x 511.18 = 3067.08, A(0) = 3000, t = 0.5.

We solve the equation for r to find the APR.

$A(t) = A(0)(1 + rt)$

$3067.08 = 3000(1 + 0.5r)$

$1 + 0.5r = \frac{3067.08}{3000}$

1 + 0.5r = 1.02236

r = (1.02236 - 1)/0.5

r = 0.04472.

More can be learned about simple interest at brainly.com/question/25296782

#SPJ1

5 0

1 year ago

3x-x-5=2(x+2) -9<br> please help

irga5000 [103]

Answer:

infinite solutions

Step-by-step explanation:

Simplify 3x-x-53x−x−5 to 2x-52x−5.

2x-5=2(x+2)-92x−5=2(x+2)−9

2 Expand.

2x-5=2x+4-92x−5=2x+4−9

3 Simplify 2x+4-92x+4−9 to 2x-52x−5.

2x-5=2x-52x−5=2x−5

4 Since both sides equal, there are infinitely many solutions.

Infinitely Many Solutions

6 0

3 years ago

Read 2 more answers

On any given day, Juan will break even if he earns as much money as he spends. On day 5, he records the expression 16 + (–16). D

Degger [83]

Yes she do break even because let me show you a expression now take a number line and then you start at -16 right and then you go down the number line and then this is how it go -16 -15 -14 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 and i just answer your Q there you go -16+16=0

7 0

3 years ago

Read 2 more answers

Match the key aspect of a function's graph with its meaning.

Margarita [4]

Answer:

Part 1) Intervals of the domain where the graph is above the x-axis (f(x) > 0)

Part 2) location on graph where input is zero (y-intercept)

Part 3) location on graph where output is zero (x-intercept)

Part 4) Intervals of the domain where the graph is below the x-axis (f(x) < 0)

Step-by-step explanation:

<u><em>Verify each case</em></u>

Part 1) we have

Intervals of the domain where the graph is above the x-axis

we know that

If the graph is above the x-axis, then the value of f(x) is positive

therefore

f(x) > 0

Part 2) we have

location on graph where input is zero

Let

x ---> the independent variable or input value

f(x) ---> the dependent variable or output value

we know that

The y-intercept is the value of f(x) (output value) when the value of x (input value) is zero

therefore

y-intercept

Part 3) we have

location on graph where output is zero

Let

x ---> the independent variable or input value

f(x) ---> the dependent variable or output value

we know that

The x-intercept is the value of x (input value) when the value of the function f(x) (output value) is zero

therefore

x-intercept

Part 4) we have

Intervals of the domain where the graph is below the x-axis

we know that

If the graph is below the x-axis, then the value of f(x) is negative

therefore

f(x) < 0

5 0

3 years ago