Ask question

Ask question

All categories

3 years ago

11

A jewelry maker purchased 20 inches of gold chain.She used 3/8 of the chain for a bracelet. How many inches of gold did she have

left?

1 answer:

Mazyrski [523]3 years ago

7 0

Answer:

12.5 inches of gold chain.

Step-by-step explanation:

To get the remaining chain of gold we need to calculate

8/8 - 3/8 = 5/8 of 20 inches

then multiply:

20 * 5/8 = 100/8

100/8 = 12.5 inches

You might be interested in

Question:4x²-9 <br>answer:

statuscvo [17]

Step-by-step explanation:

4x^2-9

=(2x)^2 -3^2

=(2x+3) (2x-3)

Hope this helps you.

3 0

2 years ago

William runs at a constant speed of 8 miles per hour. Running at this constant speed, it takes William 75 seconds to run from hi

pav-90 [236]

Answer:

0.16 miles

Step-by-step explanation:

5 0

3 years ago

What are the dangers of having overdraft protection?

Oksanka [162]

Your bank can charge you a overdraft fee, it can effect your credit score <span>if you sign up for </span>overdraft <span>protection on the account.</span>

4 0

3 years ago

Selena can make 5 pancakes in 12 minutes. which equation can be used to find m the number of minutes needed to make 20 pancakes

Tamiku [17]

I hope this helps you

5 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: <em>implicit function derivative implicit differentiation chain product rule differential integral calculus</em>

6 0

3 years ago