Since the slope is -2 because
x1: -1 x2: 0
y1: 2 y2: 0
y2-y1/x2-x1 = slope (0-2/0+1= -2/1 = -2)

You would havr to change it into point slope form so it would look like this
Y-2 = -2(x+1)
y-2 = -2x -2
+2 +2
y = -2x + 0
Hope this helps:)

8 0

3 years ago

Match each of the trigonometric expressions below with the equivalent non-trigonometric function from the following list. Enter

andreev551 [17]

Answer:

Step-by-step explanation:

In each case, draw the right triangle which produces the inverse trig value. That is, label the two sides as needed, and calculate the third side.

It should be clear that

$tan(sin^{-1} x/4) = x/\sqrt{16-x^{2} }$

$sin(tan^{-1} x/4) = x/\sqrt{16+x^{2} }$

sin(2α) = 2 sinα cosα

So,

$1/2 sin(2sin^{-1} x/4) = (1/2)(x/4) * x/\sqrt{16-x^2} = x^{2} /(8\sqrt{16-x^2})$

See what you can do with the others.

6 0

3 years ago

What number should be added to both sides of the equation to complete the square? x^2 – 6x = 5

NeTakaya

You will add $( \frac{-6}{2} )^{2}$
$(-3)^{2}$
9

5 0

3 years ago