The legend on a map states that

Regard y as the independent variable and x as the dependent variable and use implicit differentiation to find dx/dy. x5y2 − x4y

il63 [147K]

Answer:

dx/dy = (x^4 - 2x^5y - 6xy^2) / (5x^4y^2 - 4x^3y + 2y^3).

Step-by-step explanation:

x^5y^2 − x^4y + 2xy^3 = 0

Applying the Product and Chain Rules:

y^2*5x^4*dx/dy + 2y*x^5 - (y*4x^3*dx/dy + x^4) + (y^3* 2*dx/dy + 3y^2*2x) =0

Separating the terms with derivatives:

y^2*5x^4*dx/dy - y*4x^3*dx/dy + y^3* 2*dx/dy = x^4 - 2y*x^5 - 3y^2*2x

dx/dy = (x^4 - 2x^5y - 6xy^2) / (5x^4y^2 - 4x^3y + 2y^3)

7 0

3 years ago

Read 2 more answers

if the length of the longest side of the rectangle is 20cm find the shaded area to the nearest hundredth PLEASE HELPP !

djverab [1.8K]

Answer:

150pi

Step-by-step explanation:

lets just say the two circles are 10 and 10 in diameter. so, one square is 5 radius. The square is 25pi area. so two are 50pi area. the area of the rectangle is 200, so 150pi

Please tell me if I'm wrong D:

3 0

3 years ago

Using the distributive property rewrite 2x + 16 as a product of two factors

tresset_1 [31]

2(x + 8)
Factor out the 2

5 0

3 years ago

How do you determine the area under a curve in calculus using integrals or the limit definition of integrals?

RSB [31]

Answer:

Please check the explanation.

Step-by-step explanation:

Let us consider

$y = f(x)$

To find the area under the curve $y = f(x)$ between $x = a$ and $x = b$ , all we need is to integrate $y = f(x)$ between the limits of $a$ and $b$ .

For example, the area between the curve y = x² - 4 and the x-axis on an interval [2, -2] can be calculated as:

$A=\int _a^b|f\left(x\right)|dx$

= $\int _{-2}^2\left|x^2-4\right|dx$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx$

$=\int _{-2}^2x^2dx-\int _{-2}^24dx$

solving

$\int _{-2}^2x^2dx$

$\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1$

$=\left[\frac{x^{2+1}}{2+1}\right]^2_{-2}$

$=\left[\frac{x^3}{3}\right]^2_{-2}$

computing the boundaries

$=\frac{16}{3}$

Thus,

$\int _{-2}^2x^2dx=\frac{16}{3}$

similarly solving

$\int _{-2}^24dx$

$\mathrm{Integral\:of\:a\:constant}:\quad \int adx=ax$

$=\left[4x\right]^2_{-2}$

computing the boundaries

$=16$

Thus,

$\int _{-2}^24dx=16$

Therefore, the expression becomes

$A=\int _a^b|f\left(x\right)|dx=\int _{-2}^2x^2dx-\int _{-2}^24dx$

$=\frac{16}{3}-16$

$=-\frac{32}{3}$

$=-10.67$ square units

Thus, the area under a curve is -10.67 square units

The area is negative because it is below the x-axis. Please check the attached figure.

6 0

3 years ago

A rental car company charges $30.05 per day to rent a car and $0.10 for every mile driven. Anthony wants to rent a car, knowing

Annette [7]

Answer:

Three days at the most.

Step-by-step explanation:

First you must see how much money he spends on driving, 75 x 0.10 = 7.5. Subtract that from 120 to get 112.5. He has that much money left. Now you will divide 112.5 by 30.05. 112.5/30.05 = 3.74376039933. Simplify to get 3.

(also, he will have 22.35 dollars left)

7 0

3 years ago