Find the equation of the line that goes through the points (2, 2) and (6, 6).

Mathematics

2 answers:

atroni [7]2 years ago

7 0

Answer:

Step-by-step explanation:

y=6x+2

2=12+2

NO
y=-x

2=-2

y=2x+6

2=4+10

y=x

2=2

6=6

notka56 [123]2 years ago

3 0

Answer:

$y=x$

Step-by-step explanation:

We will use slope-intercept form:

$y=mx+b$

Where $m=$ slope and $b =$ y-intercept

To find the slope we use the slope formula:

$\frac{y_2-y_1}{x_2-x_1}$

We plug in the coordinates getting:

$\frac{6-2}{6-2} =\frac{4}{4} = 1$

So we have a slope of 1:

$y=(1)x+b$

If we look at the answer choice we see that only one of them has a slope of 1 so the equation is:

$y =x$

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3 years ago

Which are the solutions of x^2 = –11x + 4?

strojnjashka [21]

This is a quadratic equation, i.e. an equation involving a polynomial of degree 2. To solve them, you must rearrange them first, so that all terms are on the same side, so we get

$x^2 + 11x - 4 = 0$

i.e. now we're looking for the roots of the polynomial. To find them, we can use the following formula:

$x_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2}$

where $x_{1,2}$ is a compact way to indicate both solutions $x_1$ and $x_2$ , while $a,b,c$ are the coefficients of the quadratic equation, i.e. we consider the polynomial $ax^2+bx+c$ .

So, in your case, we have $a=1,\ \ b=11,\ \ c=-4$

Plug those values into the formula to get

$x_{1,2} = \frac{-11\pm\sqrt{121+16}}{2} = \frac{-11\pm\sqrt{137}}{2}$

So, the two solutions are

$x_1 = \frac{-11+\sqrt{137}}{2}$

$x_2 = \frac{-11-\sqrt{137}}{2}$