Marica purchase three doughnuts for 2 dollers each the cashier told her when you buy 2 donuts, you get the third one half price

Algebra 2 Math question Grade 11

Paha777 [63]

Answer:

$\displaystyle y=\frac{4}{5}x+\frac{13}{5}$

Step-by-step explanation:

The equation of the line in slope-intercept form is:

y=mx+b

Where m the slope of the line and b the y-intercept.

When two points are given, it's convenient to calculate the slope first.

Suppose we know the line passes through points A(x1,y1) and B(x2,y2). The slope can be calculated with the equation:

$\displaystyle m=\frac{y_2-y_1}{x_2-x_1}$

The points are (3,5) and (-2,1):

$\displaystyle m=\frac{1-5}{-2-3}=\frac{4}{5}$

The equation is now:

$\displaystyle y=\frac{4}{5}x+b$

To calculate b, we use any of the given points and solve for b. Use (3,5):

$\displaystyle 5=\frac{4}{5}3+b$

Operate:

$\displaystyle 5=\frac{12}{5}+b$

Solve:

$\displaystyle b=5-\frac{12}{5}=\frac{13}{5}$

The required equation of the line is:

$\boxed{\displaystyle y=\frac{4}{5}x+\frac{13}{5}}$

3 0

4 years ago

Given that y = sin(x+y),find the derivative when (x,y)=(π,0)

lisov135 [29]

<h2>Answer:</h2>

Shown in the explanation

<h2>Step-by-step explanation:</h2>

Recall that an implicit function is a relation given by the form:

$R(x_{1},\ldots, x_{n})=0$

Where $R$ is a function of two or more variables. In this case, that function is:

$y = sin(x+y)$

and is implicit because we can define it as:

$y-sin(x+y)=0$ having two variables.

So, let's take the derivative:

$\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(\sin \left(x+y\right)\right) \\ \\$

Applying chain rule:

$\frac{d}{dx}\left(\sin \left(x+y\right)\right)=\cos \left(x+y\right)\left(1+\frac{d}{dx}\left(y\right)\right)$

But:

$\frac{d}{dx}\left(y\right)=y'$

Therefore:

$y'=\cos \left(x+y\right)\left(1+y'\right)$

Isolating $y'$ :

$\frac{d}{dx}\left(y\right)=y'=\frac{\cos \left(x+y\right)}{1-\cos \left(x+y\right)}$

When $(x,y)=(\pi,0)$ :

$\frac{d}{dx}\left(y\right)|_{(\pi,0)}=\frac{\cos \left(\pi+0\right)}{1-\cos \left(\pi+0\right)} \\ \\ \frac{d}{dx}\left(y\right)|_{(\pi,0)}=\frac{\cos \left(\pi\right)}{1-\cos \left(\pi\right)} \\ \\ \frac{d}{dx}\left(y\right)|_{(\pi,0)}=\frac{-1}{1-(-1)} \\ \\ \boxed{\frac{d}{dx}\left(y\right)|_{(\pi,0)}=-\frac{1}{2}}$