The easiest way is to try the point (-4,1), that is, x=-4, y=1,
to see which equation works.
b works.

The usual way to do it is to find the equation of the circle
standard form of a circle is (x-h)²+(y-k)²=r², (h,k) are the coordinates of the center, r is the radius.
in this case, the center is (-2,1), so (x+2)²+(y-1)²=r²
the given point (-4,1) is for you to find r: (-4+2)²+(1-1)²=r², r=2
so the equation is (x+2)²+(y-1)²=2²
expand it: x²+4x+4+y²-2y+1=4
x²+y²+4x-2y+1=0, which is answer b.

5 0

3 years ago

What is the square root of 18 plus the square root of 2? And is it rational or irrational?<br>

kicyunya [14]

The answer is (rounded to the nearest tenth) about 5.7 and since neither 18 or 2 have a perfect square, so this is irrational.

6 0

2 years ago

Find the equation of a line that passes through the point (3,1) and has a gradient of -3. Leave your answer in the form y = m x

Bingel [31]

Answer:

$y = (-3)\, x + 10$ .

Step-by-step explanation:

The question requires that the equation should be in the slope-intercept form $y = m \, x + c$ . The $m$ and $c$ in this equation are constants. In particular:

$m\!$ would represent the slope of this line (also known as the gradient of this line,) whereas
$c$ would be the $y$ -intercept of this line.

The question states that the slope of this line is $(-3)$ . Therefore, $m = -3$ . Thus, the equation for this line should be in the form $y = (-3)\, x + c$ .

The value of constant $c$ could be found using the other piece of information: the point $(3,\, 1)$ is in this line. In other words, the solution $\text{$x = 3$ and $y = 1$}$ should satisfy the equation $y = (-3)\, x + c$ .

Substitute $\text{$x = 3$ and $y = 1$}$ into the equation $y = (-3)\, x + c$ and solve for $c$ :

$1 = (-3) \times 3 + c$ .

$1 = -9 + c$ .

$c = 10$ .

Hence, $y = (-3) \, x + 10$ would be the slope-intercept form of the equation of this line.

3 0

2 years ago