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alexandr402 [8]
3 years ago
9

I) Domain, ii) x-intercept, iii) y-intercept, iv) vertical asymptote, and v) horizontal asymptote of

Mathematics
1 answer:
Montano1993 [528]3 years ago
7 0

Answer:

i)

Step-by-step explanation:

i) you can't divide by 0 and when you put x=5 into the bottom part 4x-20 = 4(5)-20 = 20-20 = 0 so x cannot be 5. So the domain is x > 5 and x < 5 OR in other words, when x is not 5.

ii) substitute y=0 into the equation and solve the quadratic on top

iii) substitute x=0 into the equation and write down the value

horizontal asymptote at x=5 because we found out that x cannot be 5 in part i)

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Lapatulllka [165]

Given:

The graph of line.

To find:

The gradient of the line using rise/run method.

Solution:

We know that the gradient of a line is also known as slope.

Slope=\dfrac{Rise}{Run}

Consider the two intercepts, then rise is distance between origin and y-intercept and run is the distance between origin and the x-intercept. But rise must be negative because the value of y decreased from 3 to 0.

Rise=-3

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Now,

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3 years ago
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marta [7]

Answer:

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Step-by-step explanation:

Slope is defined as "rise over run". If you count out the units, you should be able to see that your rise is 1 and your run is 3 and is pointing downward, so your answer is -1/3.

Alternatively you could use Slope Formula: m = \frac{y2-y1}{x2-x1}

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sergeinik [125]

Answer:

a). 8(x + a)

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    Now substitute these values in the expression,

    \frac{f(x)-f(a)}{x-a} = \frac{8x^2-8a^2}{x-a}

                  = \frac{8(x^2-a^2)}{(x-a)}

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b). \frac{f(x+h)-f(x)}{h} = \frac{8(x+h)^2-8x^2}{h}

                      = \frac{8(x^2+h^2+2xh)-8x^2}{h}

                      = \frac{8x^2+8h^2+16xh-8x^2}{h}

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4 years ago
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A sales person is given a choice of two salary plans. Plan 1 is a weekly salary of 700 plus 4% commission of sales. Plan 2 is a
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Let 's' represent the amount of sales.

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Divide both sides by 0.08,

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Hence, the amount of sales is $8,750.

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