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Featured 1 month ago

First, convert

Therefore,

The perpendicular line to

We know that it passes through

Therefore, the equation of the line that passes through

Featured 1 month ago

The general slope-intercept form (for example) is

(where

Sample equations that would meet your requirement:

Featured 1 month ago

#"the initial statement is "ypropx#

#"to convert to an equation multiply by k the constant"#

#"of variation"#

#rArry=kx#

#"to find k use the given condition"#

#x=6" when "y=102#

#y=kxrArrk=y/x=102/6=17#

#"equation is " color(red)(bar(ul(|color(white)(2/2)color(black)(y=17x)color(white)(2/2)|)))#

Featured 1 month ago

We can graph this equation using a sign chart.

First, let's identify the holes, vertical and horizontal asymptotes and x and y intercepts.

**Horizontal Asymptotes**

Observing the formula, we can see the degree of the denominator is higher than the numerator (The degree is the largest exponent in a polynomial). This means there will be an asymptote at

**Vertical Asymptotes**

In order to find vertical asymptotes, we muct factor the denominator. Factoring out an

**Holes**

Holes occur when there is a zero in both the numerator and denominator that will cancel. There are none in this formula.

**X-Intercepts**

To find x-intercepts, substitute 0 for

There are no x-intercepts as the equation above would result in imaginary numbers as answers (

**Y-Intercepts**

To find y-intercepts substitute 0 for

As the denominator equals zero the equation is undefined. Therefore there are no y-intercepts.

**Sign Chart**

Using the values above, also known as critical values, we will create a sign chart. We do this by sorting the critical values from least to greatest, and surrounding them with

We will then select a value for

We will test the

We don't necessarily care about the value, just whether it is positive or negative. Since it is positive, we will mark a

We are now able to graph now that we know where the function is positive and negative and we have the asymptotes.

graph{(2x^2+1)/(x^3-x) [-10, 10, -5, 5]}

Notice how the graph is positive where we noted positive, and negative where we noted negative.

Featured 1 month ago

#"the diagonal 'splits' the rectangle into 2 right triangles"#

#"using "color(blue)"Pythagoras' theorem "#

#x^2+y^2=17^2#

#"area of rectangle "=xy=120#

#"thus we have 2 equations"#

#x^2+y^2=289to(1)#

#xy=120to(2)#

#"from equation "(2)toy=120/xto(3)#

#"substitute "y=120/x" into equation "(1)#

#rArrx^2+(120/x)^2=289#

#rArrx^2+14400/x^2=289#

#"multiply through by "x^2#

#rArrx^4+14400=289x^2#

#rArrx^4-289x^2+14400=0#

#"let "u=x^2#

#rArru^2-289u+14400=0larrcolor(blue)"in standard form"#

#"with "a=1,b=-289" and "c=14400#

#"solve for u using the "color(blue)"quadratic formula"#

#u=(289+-sqrt(83521-57600))/2=(289+-161)/2#

#rArru=(289-161)/2" or "u=(289+161)/2#

#rArru=64" or "u=225#

#u=x^2rArrx=sqrtu#

#rArrx=sqrt64=8" or "x=sqrt225=15#

#"substitute these values into equation "(3)#

#x=8toy=120/8=15#

#x=15toy=120/15=8#

#"length can be "15" or "8#

#"breadth can be " 8" or "15#

#rArr15xx8" or "8xx15" are the dimensions"#

Featured 4 weeks ago

From

and

we get

now squaring both sides

now subtracting side by side

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