$\frac{x^2+x-30}{x^2-3x+2}\\\\x^2-3x+2=0\\\\x^2-x-2x+2=0\\\\x(x-1)-2(x-1)=0\\\\(x-1)(x-2)=0\iff x-1=0\ \vee\ x-2=0\\\\x=1\ \vee\ x=2\\\\Answer:For\ x=1\ or\ x=2.$

8 0

3 years ago

Which are linear pairs? Check all that apply.

boyakko [2]

Answer:

Options B, C and D

Step-by-step explanation:

Linear pair of angles means sum of the angles of the pair is 180° which are formed at a point of a line.

In other words, pair of supplementary angles is the linear pair.

From the figure attached,

Pair of the linear angles are,

1). ∠BAC and ∠CAD

2). ∠BAE and ∠EAD

3). ∠EAD and ∠DAC

Therefore, Options B, C and D are correct.

5 0

3 years ago

Read 2 more answers

Calculate the perimeter of a right scalene triangle with a hypotenuse of 10 and a leg of 6. Enter just the number for your answe

Anarel [89]

Answer:

Step-by-step explanation:

8 0

3 years ago

Given f(x) = x^2 − 2x − 5.

lukranit [14]

Answer:

$(x - 1)^2 - 6$

Step-by-step explanation:

You need to know:

Vertex form = $a(x -h)^2 + k$

The vertex is at

(h, k)

Need to know about perfect squares

Need to know how to complete the square.

-----------------------------------------------------------------------------------------

To convert $x^2-2x-5$ you need to complete the square on the equation.

Complete the Square $x^2 - 2x -5$

$x^2 - 2x -5$

$(x^2 - 2x) -5$

Divide -2 by 2 and then square it.

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

$-1^2 = 1$

Add the one to the parentheses and subtract the one from the 5

$(x^2 - 2x +1) -5 -1$

Square $(x^2 - 2x +1)$

$(x-1)^2$

Now we have

$(x - 1)^2 -5 -1$

Next add -5 - 1 = -6

$(x - 1)^2 - 6$

Our quadratic is in vertex form now.

Vertex form = $a(x -h)^2 + k$

our equation = $(x - 1)^2 - 6$

Vertex = (1, -6)

5 0

4 years ago

If the numbers (-10,-20, 85,50) were temperatures which number would be the coldest?​

If the numbers (-10,-20, 85,50) were temperatures which number would be the coldest?