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3 years ago

Find the equation of a line parallel to y - 5x = 10 that passes through the point (3, 10). (answer in slope-intercept form)

Mathematics

2 answers:

MariettaO [177]3 years ago

8 0

Answer:

Step-by-step explanation:

NeX [460]3 years ago

6 0

So, a line parallel to y - 5x = 10, will have the same slope as that equation, so what is that slope anyway? let's solve for "y".

 $\bf y-5x=10\implies y=5x+10\implies y=\stackrel{slope}{5}x\stackrel{y-intercept}{+10}$

alrite, so the slope is 5 then, well, then the parallel line will have the same slope.

so, we're really looking for the equation of a line whose slope is 5 and runs through 3,10.

 $\bf \begin{array}{lllll}
&x_1&y_1\\
% (a,b)
&({{ 3}}\quad ,&{{ 10}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 5
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-10=5(x-3)
\\\\\\
y-10=5x-15\implies y=5x-5$

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The demand equation illustrates the price of an item and how it relates to the demand of the item.

The slope of the demand function is -1/2
The equation of the demand function is: $R(x) = (300 - 10x) \times (20 + 5x)$
The price that maximizes her revenue is: Ghc 85

From the question, we have:

$Plates = 300$

$Price = 20$

The number of plates (x) decreases by 10, while the price (y) increases by 5. The table of value is:

$\begin{array}{cccccc}x & {300} & {290} & {280} & {270} & {260} \ \\ y & {20} & {25} & {30} & {35} & {40} \ \end{array}$

The slope (m) is calculated using:

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

So, we have:

$m = \frac{25-20}{290-300}$

$m = \frac{5}{-10}$

$m = -\frac{1}{2}$

The equation of the demand is as follows:

The initial number of plates (300) decreases by 10 is represented as: (300 - 10x).

Similarly, the initial price (20) increases by 5 is represented as: (20 + 5x).

So, the demand equation is:

$R(x) = (300 - 10x) \times (20 + 5x)$

Open the brackets to calculate the maximum revenue

$R(x) =6000 + 1500x - 200x - 50x^2$

$R(x) =6000 + 1300x - 50x^2$

Equate to 0

$6000 + 1300x - 50x^2 =0$

Differentiate with respect to x

$1300 - 100x =0$

Collect like terms

$100x =1300$

Divide by 100

$x =13$

So, the price at maximum revenue is:

$Price= 20 + 5x$

$Price= 20 + 5 * 13$

$Price= 85$

In conclusion:

The slope of the demand function is -1/2
The equation of the demand function is: $R(x) = (300 - 10x) \times (20 + 5x)$
The price that maximizes her revenue is: Ghc 85