What is the slope of the line that passes through the points (2, 8) and (6, 12)?

1 answer:

5 0

Yeah, where I live, to find the slope (or as we call it, the gradient), you just have to do the two y values subtracted from each other, over the two x values subtracted from each other. The important thing is to remember that if you subtract the first y value from the second (so, 12 - 8), you have to do the same thing for the x (6 - 2)

$\frac{12-8}{6-2}$ = $\frac{4}{4}$ = 1.

The slope is 1.

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1.) Use the quadratic formula to solve the equation. If necessary, round to the nearest hundredth .x squared minus 21 x equals n

Andrew [12]

Part 1
We are given $x^2-21x=-4x$ . This can be rewritten as $x^2-18x=0$ .
Therefore, a=1, b=-18, c=0.
Using the quadratic formula
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-\left(-18\right)\pm \sqrt{\left(-18\right)^2-4\left(1\right)\left(0\right)}}{2\left(1\right)}$
$x=\frac{18\pm 18}{2}$

The values of x are
$x_1=\frac{18-18}{2}=0$
$x_2=\frac{18+18}{2}=18$

Part 2
Since the values of y change drastically for every equal interval of x, the function cannot be linear. Therefore, the kind of function that best suits the given pairs is a quadratic function.

Part 3.
The first equation is $y=x^2+2$ .
The second equation is $y=3x+20$ .

We have
$x^2+2=3x+20$
$x^2-3x-18=0$
Factoring, we have
$\left(x-6\right)\left(x+3\right)=0$
Equating both factors to zero.
$x_1-6=0\rightarrow x_1=6$
$x_2+3=0\rightarrow x_2=-3$

When the value of x is 6, the value of y is
$y=3\left(6\right)+20=38$

When the value of x is -3, the value of y is
$y=3\left(-3\right)+20=11$

Therefore, the solutions are (6,38) or (-3,11)

7 0

2 years ago

Graph the following features:<br> Y-intercept = -2<br> Slope = -3

SVEN [57.7K]

Slope intercept formula
Y=-3x-2
Plot a point at the intercept (0,-2)
Then go down 3 over 1 and plot another point (1,-5) then connect with a line