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3 years ago

PLEASE HELPP!! DID I DO IT RIGHT???

Mathematics

1 answer:

astra-53 [7]3 years ago

8 0

Answer:

yea haha good job :-)

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Which pair of lines are perpendicular

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3 years ago

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Factor the following and find the zeros.

yaroslaw [1]

$1.\ x^2-169=0\\\\x^2-13^2=0\\\\(x-13)(x+13)=0\iff x=13\ \vee\ x=-13$

$2.\ x^2+14x+49=0\\\\x^2+2x\cdot7+7^2=0\\\\(x+7)^2=0\\\\(x+7)(x+7)=0\iff x=-7$

$3.\ x^2-17x+42=0\\\\x^2-14x-3x+42=0\\\\x(x-14)-3(x-14)=0\\\\(x-14)(x-3)=0\iff x=14\ \vee\ x=3$

$4.\ 2x^2-9x-5=0\\\\2x^2-10x+x-5=0\\\\2x(x-5)+1(x-5)=0\\\\(x-5)(2x+1)=0\iffx=5\ \vee\ x=-\frac{1}{2}$

$5.\ 3x^2+11x-4=0\\\\3x^2+12x-x-4=0\\\\3x(x+4)-1(x+4)=0\\\\(x+4)(3x-1)=0\iff x=-4\ \vee\ x=\frac{1}{3}$

$6.\ this\ same\ like\ 1.\\\\x^2-196=0\\\\x^2-14^2=0\\\\(x-14)(x+14)=0\iff x=14\ \vee\ x=-14$

3 0

3 years ago

How to solve this problem?

valentinak56 [21]

Very nice handwriting but the math and English are confusing.

Let's assume we're told

$\displaystyle 45 = \sum_{i=1}^9 (x_i - 10)^2$

The subscript is important.

I think we're told the similar sum with 11 gives the smallest possible value for the sum. This is a rather cagey way of telling us 11 is the mean of the nine points. The mean is the number which minimizes the sum of squared deviations.

$\displaystyle 45 = \sum_{i=1}^9 (x_i - 10)^2 = \sum x_i^2 - 20 \sum x_i + 9(100)$

$\displaystyle \sum x_i^2= 20 \sum x_i - 900$

If 11 is the mean, the sum of the points is 9(11)=99.

$\displaystyle \sum x_i^2= 20 (99) - 900 = 1080$

Answer: 1080

6 0

4 years ago