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3 years ago

Transform the quadratic function defined by y=ax2+bx+c into the form y=a(x-h)2+k 1.Y=x2-6x-3 2.Y=5x2-20x-5

Mathematics

1 answer:

Lana71 [14]3 years ago

7 0

Answer:

1) y = (x-3)²-12

2) y = 5(x-2)²-9 

Step-by-step explanation:

Given the quadratic expressions

y=x²-6x-3

We are to transform it in the form y=ax²+bx+c

y=(x²-6x)-3

Complete the square of the expression in parenthesis

y = (x²-6x + (6/2)² - (6/2)²)-3

y = (x²-6x+3²-3²)-3

y = (x²-6x+9-9)-3

y = (x²-6x+9)-12

y = (x²-3x-3x+9)-12

y = x(x-3)-3(x-3)-12

y = (x-3)(x-3)-12

y = (x-3)²-12

Hence the transformation is y = (x-3)²-12 where a = 1 and k = -12

For the quadratic equation

y=(5x²-20x)-5

y = 5(x²-4x)- 5

Complete the square of the expression in parenthesis

y = 5(x²-4x+4-4)- 5

y = 5(x²-4x+4)-9

y = 5(x-2)²-9

Hence the transformation is y = 5(x-2)²-9 where a = 5 and k = -9

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3 years ago

the length of a rectangle is increasing ata a ate of 8cm and is width is increasing at a rate of 3cm when the length is 20cm and

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Answer:

us set up the following variables:

⎧

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⎨

⎪

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3 years ago

Using the pattern, give the coefficients of (x + y)^5 and (x + y)^6

musickatia [10]

Answer:

$(x+y)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

$(x+y)^6=x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6$

Step-by-step explanation:

In order to find the values of $(x+y)^5$ and $(x+y)^6$ , you need to apply the binomial theorem (high-level math you most likely don't need to worry about, it's easier than multiplying all the binomials together).

$(x+y)^5$ = $\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$ = $\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$ = $x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$ .

$(x+y)^6$ = $\sum _{i=0}^6\binom{6}{i}x^{\left(6-i\right)}y^i$ = $\frac{6!}{0!\left(6-0\right)!}x^6y^0+\frac{6!}{1!\left(6-1\right)!}x^5y^1+\frac{6!}{2!\left(6-2\right)!}x^4y^2+\frac{6!}{3!\left(6-3\right)!}x^3y^3+\frac{6!}{4!\left(6-4\right)!}x^2y^4+\frac{6!}{5!\left(6-5\right)!}x^1y^5+\frac{6!}{6!\left(6-6\right)!}x^0y^6$ = $x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6$ .

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