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soldi70 [24.7K]
3 years ago
5

Factor each. Please help!

Mathematics
1 answer:
Naddika [18.5K]3 years ago
7 0

Answer:

1) The factors of  x^2-3x+2=0 are \mathbf{(x-1)(x-2)=0}

Option C is correct.

3) The factors of  x^2+2x-8=0 are \mathbf{(x-2)(x+4)=0}

Option B is correct.

Step-by-step explanation:

1) Factor : x^2-3x+2=0

For factoring we need to break the middle term, such that there sum is equal to middle term of expression and product is equal to product of first and last term.

The middle term : -3x

We can break them as (-2x)( -x)

Solving:

x^2-3x+2=0\\x^2-2x-x+2=0\\x(x-2)-1(x-2)=0\\(x-1)(x-2)=0

So, factors of  x^2-3x+2=0 are \mathbf{(x-1)(x-2)=0}

Option C is correct.

3) Factor: x^2+2x-8=0

For factoring we need to break the middle term, such that there sum is equal to middle term of expression and product is equal to product of first and last term.

The middle term : 2x

We can break them as (4x)( -2x)

Solving

x^2+2x-8=0\\x^2+4x-2x-8=0\\x(x+4)-2(x+4)=0\\(x-2)(x+4)=0

So, factors of  x^2+2x-8=0 are \mathbf{(x-2)(x+4)=0}

Option B is correct.

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4 years ago
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3 years ago
Work out the formula of the nth term in the following quadratic sequence:<br> 19, 15, 9, 1...
Butoxors [25]

In a quadratic sequence we'll get a linear first difference and a constant second difference.  Let's verify that.

n          1   2   3   4

f(n)       19  15  9   1

1st diff   -4  -6  -8

2nd diff     2   2

We see that we got a constant second difference.  We could just extend that and work back up to get more values.

n          1   2   3   4       5    6      7

f(n)       19  15  9   1      -9   -21   -35

1st diff   -4  -6  -8   -10  -12   -14

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That's just an aside; we're after the general formula.  We have

f(1)=19, f(2)=15, f(3)=9

In general we can assume

f(n) = an²  + bn + c

We get three equations in three unknowns,

19 = a(1²)+b(1)+c = a+b+c

15 = a(2²) + b(2) + c = 4a + 2b + c

9 = a(3²) + b(3) + c =  9a + 3b + c

That's a 3x3 linear system; it's easy to solve directly.  Subtracting pairs,

4 = -3a - b

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Subtracting those,

-2 = 2a

a = -1

b = -3a -4 = -1

c = 19-a-b = 21

Answer: f(n) = -n² - n + 21

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f(2) = -4 - 2 + 21 = 15, good

f(3) = -9 - 3 + 21 = 9, good

f(4) = -16 - 4 + 21 = 1, good

Let's check our extended table, how about

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Answer:

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Step-by-step explanation:

given,

mean of five distinct positive number = 1000

median of the number = 100

100 is median means two number will be less than 100 and two number will be greater than 100.

let five number be

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'e' should be the largest number

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'a' and 'b' should be as small as possible and d should be the number nearest to 100.

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now,

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1 + 2 + 100 + 101 + e = 5000

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