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3 years ago

100 points Use the line tool to graph the equation on the coordinate plane. y=3/4x

Mathematics

1 answer:

Ad libitum [116K]3 years ago

5 0

Answer:

The equation of line passes through the origin with slope m is given by:

$y=mx$

Given the equation:

$y = \frac{3}{4}x$

Slope = 3/4

y-intercept = 0

To Graph the equation on the coordinate plane.

Let us consider some values of x and y to make table.

x $y = \frac{3}{4}x$

0 0

2 1.5

4 3

6 4.5

8 6

10 7.5

Now, you can plot these points on the coordinate plane using line tool as shown below.

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A farmer wants to fence in a rectangular plot of land adjacent to the north wall of his barn. no fencing is needed along the bar

jeyben [28]

Hi,

Let assume a the west side,

b the length of the north wall (i suppose the answer is smaller then the barn.

Cost of the fence= a*6+a*12+b*12=3000 $.

So 3a+2b=500==>b=(500-3a)/2

Area =a*b=a(500-3a)/2= 250a-3/2a²

Derivate the area ==> 250-3a=0

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Width=125 (feet)

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2 years ago

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Answer:

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Step-by-step explanation:

expand he equation

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4 substituted as X eliminates all an results to zero

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3 years ago

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1 year ago

Read 2 more answers

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>

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2 years ago