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slega [8]
3 years ago
14

50 over 8 as a reduced fraction

Mathematics
2 answers:
Elenna [48]3 years ago
4 0
\frac{50}{8} as a reduced fraction is \frac{25}{4}

To figure out this answer we can simply look at our original fraction. \frac{50}{8}, and see what the GCF or greatest common factor of the two numbers are.

When we're talking about GCF, we are looking for the largest number that is divisible by every number we are looking at. That being said, what is the largest number that can be divided into both 50 and 8. The number that we find is 2.

Now that we have our GCF, all we have to do is divide both the numerator and the denominator by 2:

\frac{50}{8}÷ 2 = \frac{25}{4}

If we notice, there is no number that goes into 25 that can also go into 4, so we have our fraction as reduced as possible, or in simplest terms.

So, our final answer is: \frac{25}{4}

Thank you for your question! I hope this helped! :) Have an amazing day! :D


BartSMP [9]3 years ago
3 0
This Is The Answer Hope It Helps

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Solve the following quadratics. State the FACTORS AND SOLUTIONS. 1. 2x^2 - 7x + 3 2. 3x^2 + 7x +2
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1. x = 3, 1/2 (solutions); (x - 3)(2x - 1) (factors)

2. x = -1/3, -2 (solutions); (3x + 1)(x + 2) (factors)

Step-by-step explanation:

<u>1. 2x^2 - 7x + 3</u>

To solve problem 1, you will need to identify your a, b, and c values in this quadratic function.

Since this problem is in standard form, it will be easy to identify these values. The standard form of a quadratic function is ax^2 + bx + c.

The a value is 2, the b value is -7, and the c value is 3 if we use our standard form and see which numbers are plugged into it.

Since we know that

  • a = 2
  • b = -7
  • c = 3

we can use the quadratic formula: x = \frac{-b~\pm~\sqrt{b^2~-~4ac} }{2a}

Substitute the a, b, and c values into the quadratic formula: x=\frac{-(-7)\pm\sqrt{(-7)^2-4(2)(3)} }{2(2)}

Now simplify using the laws of pemdas: x=\frac{7\pm\sqrt{(49)-(24)} }{4}

Simplify even further: x=\frac{7\pm\sqrt{(25)} }{4} \rightarrow x=\frac{7\pm (5) }{4}

Now split this equation into two equations to solve for x: x=\frac{12 }{4} ~~and~~ x=\frac{2 }{4}

12/4 can be simplified to 3, and 2/4 can be simplified to 1/2.

This means your solutions to problem 1 is 3, 1/2.

\boxed {x=3,\frac{1}{2} }

There is also another way to solve for the quadratic functions, and this was by factoring.

If you factor 2x^2 - 7x + 3 using the bottoms-up method, you will get (x - 3)(2x - 1).

After factoring, solving for the solutions is simple because all you have to do is set each factor to 0.

  • x - 3 = 0
  • 2x - 1 = 0

After solving for x by adding 3 to both sides, or by adding 1 to both sides then dividing by 2, you will end up with the same solutions: x = 3 and x = 1/2.

<u>2. 3x^2 + 7x + 2</u>

To save time I'll be using the bottoms-up factoring method, but remember to refer back to problem 1 (quadratic formula) if you prefer that method.

Factor this quadratic function using the bottoms-up method. After factoring you will have (3x + 1)(x + 2). These are your factors.

Now to solve for x and find the solutions of the quadratic function, you will set both factors equal to 0.

  • 3x + 1 = 0
  • x + 2 = 0

Solve.

<u>First factor:</u> 3x + 1 = 0

Subtract 1 from both sides.

3x = -1

Divide both sides by 3.

x = -1/3

<u>Second factor:</u> x + 2 = 0

Subtract 2 from both sides.

x = -2

Your solutions are x = -1/3 and x = -2.

\boxed {x = -\frac{1}{3} , -2}

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