What is 20 % of 18.5 ?

PLEASE HELP ME I NEED TO PAST THIS TEST. Jack invests $30,000 with an interest rate of 4.5% compounded quarterly. After 6 years,

zavuch27 [327]

Answer:

1350

Step-by-step explanation:

Hope this helps!

3 0

2 years ago

Ken took 64 minutes to walk 4 miles on flat terrain. If he walked at a constant rate, how long would it take Ken to complete 5 m

Svetlanka [38]

Answer: 80 minutes

Step-by-step explanation:in the picture

7 0

2 years ago

Consider the function f(x) = x3 + 7. What is the inverse of f(x)?

Brrunno [24]

Answer:

f(x)=x^3+7

y=x^3+7

x=y^3+7

y^3=x-7

y=cube root (x-7)

f(x) inverse=cube root of x-7

Step-by-step explanation:

7 0

2 years ago

Given the arrangement,

Law Incorporation [45]

Answer:

the flooor length 360 and the floor breadth 4x90=360

Step-by-step explanation:

3 0

3 years ago

If x^2y-3x=y^3-3, then at the point (-1,2), (dy/dx)?

zavuch27 [327]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2866883

_______________

 dy
Find —— for an implicit function:
 dx

x²y – 3x = y³ – 3

First, differentiate implicitly both sides with respect to x. Keep in mind that y is not just a variable, but it is also a function of x, so you have to use the chain rule there:

$\mathsf{\dfrac{d}{dx}(x^2 y-3x)=\dfrac{d}{dx}(y^3-3)}\\\\\\
\mathsf{\dfrac{d}{dx}(x^2 y)-3\,\dfrac{d}{dx}(x)=\dfrac{d}{dx}(y^3)-\dfrac{d}{dx}(3)}$

Applying the product rule for the first term at the left-hand side:

$\mathsf{\left[\dfrac{d}{dx}(x^2)\cdot y+x^2\cdot \dfrac{d}{dx}(y)\right]-3\cdot 1=3y^2\cdot \dfrac{dy}{dx}-0}\\\\\\
\mathsf{\left[2x\cdot y+x^2\cdot \dfrac{dy}{dx}\right]-3=3y^2\cdot \dfrac{dy}{dx}}$

 dy
Now, isolate —— in the equation above:
 dx

$\mathsf{2xy+x^2\cdot \dfrac{dy}{dx}-3=3y^2\cdot \dfrac{dy}{dx}}\\\\\\
\mathsf{2xy+x^2\cdot \dfrac{dy}{dx}-3-3y^2\cdot \dfrac{dy}{dx}=0}\\\\\\
\mathsf{x^2\cdot \dfrac{dy}{dx}-3y^2\cdot \dfrac{dy}{dx}=-\,2xy+3}\\\\\\
\mathsf{(x^2-3y^2)\cdot \dfrac{dy}{dx}=-\,2xy+3}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{-\,2xy+3}{x^2-3y^2}\qquad\quad for~~x^2-3y^2\ne 0}$

Compute the derivative value at the point (– 1, 2):

x = – 1 and y = 2

$\mathsf{\left.\dfrac{dy}{dx}\right|_{(-1,\,2)}=\dfrac{-\,2\cdot (-1)\cdot 2+3}{(-1)^2-3\cdot 2^2}}\\\\\\
\mathsf{\left.\dfrac{dy}{dx}\right|_{(-1,\,2)}=\dfrac{4+3}{1-12}}\\\\\\
\mathsf{\left.\dfrac{dy}{dx}\right|_{(-1,\,2)}=\dfrac{7}{-11}}\\\\\\\\ \therefore~~\mathsf{\left.\dfrac{dy}{dx}\right|_{(-1,\,2)}=-\,\dfrac{7}{11}}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

Tags: implicit function derivative implicit differentiation chain product rule differential integral calculus

6 0

2 years ago