What is the slope of (-8,-3) and (4,-6)

Mathematics

1 answer:

jekas [21]3 years ago

8 0

The slope is -1/4.

Assuming you know the rise over run method, at point (-8, -3) you go down once (-1) and to the right 4 (+4) to get to the next point on the line. When you convert it to a fraction, the slope is -1/4.

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Solve the equation using the quadratic formula : x^2-10x+25=18

Sever21 [200]

$x^2-10x+25=18 \\ \\x^2-10x+25-18 = 0\\ \\ x^2-10x+ 7 =0 \\ \\a=1 , b = -10 , c= 7 \\ \\ \Delta = b^{2}-4ac = (-10)^{2}-4*1*7=100-28 = 72 \\ \\\sqrt{\Delta }=\sqrt{72}= \sqrt{2*36} =\sqrt{36}*\sqrt{2} =6\sqrt{2}$

$x_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{10- 6\sqrt{2}}{2}=\frac{2(5-3\sqrt{2})}{2}= 5-3\sqrt{2}\\ \\x_{2}=\frac{-b+\sqrt{\Delta }}{2a} =\frac{10+ 6\sqrt{2}}{2}=\frac{2(5+3\sqrt{2})}{2}= 5+3\sqrt{2}$

$Answer : \ x= 5-3\sqrt{2} \ \ or \ \ x = 5+3\sqrt{2}$

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3 years ago

In a rational function, is the horizontal shift represented by the vertical asymptote?

ziro4ka [17]

<h3>Short Answer: Yes, the horizontal shift is represented by the vertical asymptote</h3>

A bit of further explanation:

The parent function is y = 1/x which is a hyperbola that has a vertical asymptote overlapping the y axis perfectly. Its vertical asymptote is x = 0 as we cannot divide by zero. If x = 0 then 1/0 is undefined.

Shifting the function h units to the right (h is some positive number), then we end up with 1/(x-h) and we see that x = h leads to the denominator being zero. So the vertical asymptote is x = h

For example, if we shifted the parent function 2 units to the right then we have 1/x turn into 1/(x-2). The vertical asymptote goes from x = 0 to x = 2. This shows how the vertical asymptote is very closely related to the horizontal shifting.

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2 years ago

DO THIS FOR 100 POINTS AND MARKED AS BRAINLIEST

Allushta [10]

Convert to common denominator

3/4 = 12/16

Multiply by 16

(12/16)/(3/16)

12/3

Your answer is 4.

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