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Lady bird [3.3K]
2 years ago
8

The gradient of the line segment between the points (2,-3) and (4,-7).

Mathematics
2 answers:
timurjin [86]2 years ago
4 0

Answer:

gradient = - 2

Step-by-step explanation:

Calculate the gradient (slope ) m using the gradient formula

m = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

with (x₁, y₁ ) = (2, - 3 ) and (x₂, y₂ ) = (4, - 7 )

m = \frac{-7-(-3)}{4-2} = \frac{-7+3}{2} = \frac{-4}{2} = - 2

skad [1K]2 years ago
3 0

To find the gradient of a line you use this equation: Rise / Run

I am assuming this is a graph where both the x and y-axis increase in value by one.

So first of all, you should draw out this graph.

Second, draw a point at each of the given coordinates.

Now, join these points by drawing a right angle triangle. Put simply, draw a line from the point (4, -7) down until it is on the same level as the point (2, -3), then draw a line across.

Finally, measure the length of both these sides and use them in the equation above.

Let's assume the rise (vertical line) and the run (horizontal line) are 5 and 8 respectively. We can do 5/8 to get a gradient which is 0.625.

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\mathrm{Use\:the\:rational\:root\:theorem}

a_0=15,\:\quad a_n=2

\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:3,\:5,\:15,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1,\:2

\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:3,\:5,\:15}{1,\:2}

-\frac{3}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+3

-\frac{3}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+3

\mathrm{Compute\:}\frac{2x^3-5x^2-28x+15}{x+3}\mathrm{\:to\:get\:the\:rest\:of\:the\:eqution:\quad }2x^2-11x+5

=\left(x+3\right)\left(2x^2-11x+5\right)

Factor: 2x^2-11x+5

2x^2-11x+5=\left(2x^2-x\right)+\left(-10x+5\right)

=x\left(2x-1\right)-5\left(2x-1\right)

2x^3-5x^2-28x+15=\left(x+3\right)\left(x-5\right)\left(2x-1\right)

\left(x+3\right)\left(x-5\right)\left(2x-1\right)=0

\mathrm{Using\:the\:Zero\:Factor\:Principle:}

thus zeros of f(x) is

x=-3,\:x=5,\:x=\frac{1}{2}

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3 years ago
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