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ira [324]
3 years ago
5

Please I need answers Factorize x²-6x+9

Mathematics
1 answer:
miskamm [114]3 years ago
4 0

Answer:

The factors of the given equation are:

x^2-6x+9=(x-3)(x-3)=(x-3)^2

Step-by-step explanation:

We have :

x^2-6x+9

Using middle term splitting theorem to factorize the expression :

x^2-3x-3x+9

x(x-3)+(-3)(x-3)

(x-3)(x-3)

x^2-6x+9=(x-3)(x-3)=(x-3)^2

The factors of the given equation are:

x^2-6x+9=(x-3)(x-3)=(x-3)^2

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Sum of all digits in integers from one and a billion.
Daniel [21]

Answer:

The is coming this:-

500,000,500,000

Did you get this in vedantu

8 0
3 years ago
How do u get this?? I know the answer but then I don't know why they shuld divide by 2 I will mark the person who explains wit p
Greeley [361]

Answer:

10.89

Step-by-step explanation:

Circle:

C = TTD

C = TT(6.6)

C = 20.73451151

C = 20.73

Triangle:

A = bh/2

A = 6.6(3.3)/2        6.6/2=3.3

A = 21.78/2

A = 10.89

Shaded Region = 10.89

5 0
3 years ago
HURRY NEED HELP ASAP
poizon [28]
I=p*r*t
The answer is 1131.20
5 0
3 years ago
Read 2 more answers
When playing American​ roulette, the croupier​ (attendant) spins a marble that lands in one of the 38 slots in a revolving turnt
neonofarm [45]

Answer:

Check the explanation

Step-by-step explanation:

1) probability for odd or 00 = (18+1)/38 = 0.5

2) expected value can be calculated as follows :

14 * win probability - 13 * lose probability

=14 * (18/38) - 13*(20/38)

= -0.21

7 0
3 years ago
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How to find imaginary zeros anf real zeros of F(x)=-4x^5-8x^3+12x​
Fofino [41]

Answer:

x=\{0, -1, 1, -i\sqrt{3}, i\sqrt{3}\}

Step-by-step explanation:

We are given the function:

f(x)=-4x^5-8x^3+12x

And we want to finds its zeros.

Therefore:

0=-4x^5-8x^3+12x

Firstly, we can divide everything by -4:

0=x^5+2x^3-3x

Factor out an x:

0=x(x^4+2x^2-3)

This is in quadratic form. For simplicity, we can let:

u=x^2

Then by substitution:

0=x(u^2+2u-3)

Factor:

0=x(u+3)(u-1)

Substitute back:

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

By the Zero Product Property:

x=0\text{ and } x^2+3=0\text{ and } x^2-1=0

Solving for each case:

x=0\text{ and } x=\pm\sqrt{-3}\text{ and } x=\pm\sqrt{1}

Therefore, our real and complex zeros are:

x=\{0, -1, 1, -i\sqrt{3}, i\sqrt{3}\}

4 0
3 years ago
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