Select all ratios equivalent to 12:6.

13. Write

Bingel [31]

First off, we factor out the expression:

$\displaystyle \large{y = 2 {x}^{2} - 12x + 16} \\ \displaystyle \large{y = 2 ( {x}^{2} - 6x + 8) }$

In the bracket, separate 8 out of the expression.

$\displaystyle \large{y = 2[ ( {x}^{2} - 6x + 8)] }\\ \displaystyle \large{y = 2[ ( {x}^{2} - 6x) + 8]}$

In x^2-6x, find the third term that can make up or convert it to a perfect square form. The third term is 9 because:

$\displaystyle \large{ {(x - 3)}^{2} = {x}^{2} - 6x + 9}$

So we add +9 in x^2-6x.

$\displaystyle \large{y = 2[ ( {x}^{2} - 6x + 9) + 8]}$

Convert the expression in the small bracket to perfect square.

$\displaystyle \large{y = 2[ {(x - 3)}^{2} + 8]}$

Since we add +9 in the small bracket, we have to subtract 8 with 9 as well.

$\displaystyle \large{y = 2[ {(x - 3)}^{2} + 8 - 9]} \\ \displaystyle \large{y = 2[ {(x - 3)}^{2} - 1]}$

Then we distribute 2 in.

$\displaystyle \large{y = 2[ {(x - 3)}^{2} - 1]} \\$

$\displaystyle \large{y = 2[ {(x - 3)}^{2} - 1]} \\ \displaystyle \large{y = [2 \times {(x - 3)}^{2} ]+[ 2 \times ( - 1)] } \\ \displaystyle \large{y = 2 {(x - 3)}^{2} - 2 }$

Remember that negative multiply positive = negative.

Hence the vertex form is y = 2(x-3)^2-2 or first choice.

4 0

3 years ago

The triangles are similar. What is "a" equal to? What is "b" equal to?

TEA [102]

Yes they are if that helps, if you still are unsure than just mesure them and you will see

5 0

3 years ago

Read 2 more answers

Here are the monthly charges for jo’s Mobile phone. Monthly charge £16,150 free minutes,then 13p per minute,150 free texts,then

zhuklara [117]

Answer:

16,652

Step-by-step explanation:

5 0

4 years ago

The graph of which function passes through (0,3) and has an amplitude of 3? f (x) = sine (x) + 3 f (x) = cosine (x) + 3 f (x) =

Cloud [144]

Answer:

$f(x)=3*cosine(x)$

Step-by-step explanation:

We are looking for a trigonometric function which contains the point (0, 3), and has an amplitude of 3.

We know that for a sine function $f(x)=sin(x)$ , $f(0)= 0$ ; therefore the function we a looking for cannot be a sine function because it is zero at $x=0$ .