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disa [49]
3 years ago
12

Please help with this.I need more comparing parts and all of the corresponding parts.

Mathematics
1 answer:
Andrew [12]3 years ago
3 0
Not sure about the statement
movement down 5 units
move right 2 units
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Rewrite this standard form quadratic equation into vertex form by completing the square. Upload your work here to show your thin
Alex787 [66]

Solution :

Given the equation :

$y = x^2 - 6x+3 $

$y=1(x^2 - 6x +n)+3$

$n = \left(\frac{b}{2}\right)^2$

$n = \left(\frac{6}{2}\right)^2$

n = 9

$y=1(x^2 - 6x +9-9)+3$

$y=1(x^2 - 6x +9)-9+3$

$y=1(x-3)^2- 6$

Therefore the vertex form of $y = x^2 - 6x+3 $  is  $y=1(x-3)^2- 6$.

3 0
2 years ago
If a,b,c are all non-zero numbers and a+b+c=0, prove that
Butoxors [25]

<em>Given - a+b+c = 0</em>

<em>To prove that- </em>

<em>a²/bc + b²/ac + c²/ab = 3</em>

<em>Now we know that</em>

<em>when x+y+z = 0,</em>

<em>then x³+y³+z³ = 3xyz</em>

<em>that means</em>

<em> (x³+y³+z³)/xyz = 3 ---- eq 1)</em>

<em>Lets solve for LHS</em>

<em>LHS = a²/bc + b²/ac + c²/ab</em>

<em>we can write it as LHS = a³/abc + b³/abc + c</em><em>³</em><em>/abc</em>

<em>by multiplying missing denominators,</em>

<em>now take common abc from denominator and you'll get,</em>

<em>LHS = (a³+b³+c³)/abc --- eq (2)</em>

<em>Comparing one and two we can say that</em>

<em>(a³+b³+c³)/abc = 3</em>

<em>Hence proved,</em>

<em>a²/bc + b²/ac + c²/ab = 3</em>

6 0
2 years ago
Help please I’m tired
klasskru [66]

Answer:

8.11

Step-by-step explanation:

7.14 right answer should be

8 0
2 years ago
What does this equal<br> 2 [30/5] =
lorasvet [3.4K]
The correct answer is 12
6 0
3 years ago
Read 2 more answers
Pls help me for brainliest answer answer needs to be correct
ira [324]

Answer:

a) 17, 29, 41, 53, 65....

b) 12n + 5

3 0
3 years ago
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