x + y = 3
- x - x
y = -x + 3

y - y₁ = m(x - x₁)
y - (-1) = -1[x - (-1)]
y + 1 = -1(x + 1)
y + 1 = -1(x) - 1(1)
y + 1 = -x - 1
- 1 - 1
y = -x - 2

The answer is B.

6 0

4 years ago

Read 2 more answers

Prove that :( 1 + 1/<img src="https://tex.z-dn.net/?f=tan%5E%7B2%7DA" id="TexFormula1" title="tan^{2}A" alt="tan^{2}A" align="ab

Aliun [14]

Answer:

See explanation

Step-by-step explanation:

Simplify left and right parts separately.

Left part:

$\left(1+\dfrac{1}{\tan^2A}\right)\left(1+\dfrac{1}{\cot ^2A}\right)\\ \\=\left(1+\dfrac{1}{\frac{\sin^2A}{\cos^2A}}\right)\left(1+\dfrac{1}{\frac{\cos^2A}{\sin^2A}}\right)\\ \\=\left(1+\dfrac{\cos^2A}{\sin^2A}\right)\left(1+\dfrac{\sin^2A}{\cos^2A}\right)\\ \\=\dfrac{\sin^2A+\cos^2A}{\sin^2A}\cdot \dfrac{\cos^2A+\sin^A}{\cos^2A}\\ \\=\dfrac{1}{\sin^2A}\cdot \dfrac{1}{\cos^2A}\\ \\=\dfrac{1}{\sin^2A\cos^2A}$

Right part:

$\dfrac{1}{\sin^2A-\sin^4A}\\ \\=\dfrac{1}{\sin^2A(1-\sin^2A)}\\ \\=\dfrac{1}{\sin^2A\cos^2A}$

Since simplified left and right parts are the same, then the equality is true.

3 0

3 years ago