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Dmitrij [34]
3 years ago
5

X^2-25=0 Using factoring method

Mathematics
2 answers:
Tema [17]3 years ago
6 0
Hello.

Move <span><span>−25</span><span>-25</span></span> to the right side of the equation by subtracting <span><span>−25</span><span>-25</span></span> from both sides of the equation.<span><span><span>x2</span>=25</span><span><span>x2</span>=25

</span></span>Take the square root of both sides of the equation to eliminate the exponent on the left side.
<span><span>
</span></span>The complete solution is the result of both the positive and negative portions of the solution.

Simplify the right side of the equation.

Rewrite <span>2525</span> as <span><span>52</span><span>52</span></span>.

Pull terms out from under the radical, assuming positive real numbers.The complete solution is the result of both the positive and negative portions of the solution.
<span><span>
</span></span>The complete solution is the result of both the positive and negative portions of the solution.

<span>Answer: x=5,−<span>5

Have a nice day</span></span>
Alexxandr [17]3 years ago
5 0
x^2  - 25 = 0

That's the difference of two squares, which has a general factorization 

a^2 - b^2 = (a+b)(a-b)

Applying it to our case

0 = x^2 - 25 = (x+5)(x-5)

Two factors multiply to zero; either must be zero.

x=-5 \textrm{ or } x=5



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In a lab experiment, a student is trying to apply the conservation of momentum. Two identical balls, each with a mass of 1.0 kg.
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Answer:

Trial- 2 shows the conservation of momentum in a closed system.

Step-by-step explanation:

Given: Mass of balls are m= 1.0\ kg

Conservation of momentum in a closed system occurs when momentum before collision is equal to momentum after collision.

  • Let initial velocity of ball A\ is\ u_1
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  • Final velocity of ball B\ is\ v_2
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Now, According to conservation of momentum.

Momentum before collision = Momentum after collision

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We will plug each trial to this equation.

Trial 1

mu_1+mu_2=mv_1+mv_2\\1.0(1)+1.0(-2)=1.0(-2)+1.0(-1)\\1-2=-2-1\\-1=-3

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mu_1+mu_2=mv_1+mv_2\\1.0(.5)+1.0(-1.5)=1.0(-.5)+1.0(-\.5)\\.5-1.5=-.5-.5\\-1=-1

Trial 3

mu_1+mu_2=mv_1+mv_2\\1.0(2)+1.0(1)=1.0(1)+1.0(-2)\\2+1=1-2\\3=-1

Trial 4

mu_1+mu_2=mv_1+mv_2\\1.0(.5)+1.0(-1)=1.0(1.5)+1.0(-1.5)\\.5-1=1.5-1.5\\-.5=0

We can see only Trial 2 satisfies the princple of conservation of momentum. That is momentum before collison should equal to momentum after collision.

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3 years ago
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