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a_sh-v [17]
3 years ago
7

Can someone please help Simple the square root 24+ square root 25

Mathematics
2 answers:
yanalaym [24]3 years ago
8 0

Answer:  5 + 2\sqrt{6}

===========================================

Work Shown:

\sqrt{24}+\sqrt{25}\\\\\sqrt{25}+\sqrt{24}\\\\\sqrt{25}+\sqrt{4*6}\\\\\sqrt{25}+\sqrt{4}*\sqrt{6} \ \ \text{see note 1}\\\\\sqrt{5^2}+\sqrt{2^2}*\sqrt{6}\\\\5+2\sqrt{6} \ \ \text{see note 2}\\\\

----------

  • Note 1:  I used the rule sqrt(x*y) = sqrt(x)*sqrt(y)
  • Note 2: The rule sqrt(x^2) = x was used, where x is nonnegative
earnstyle [38]3 years ago
4 0

Answer:

5+(2)*(sq root 6)

Step-by-step explanation:

This is pretty simple, a given is that the square root of 25 is 5, so you already have one of your numbers. Since the square root of 24 is not a perfect square, it must be reduced and then added separately to 5. The sq root of 24 is reduced to 2 multiplied by the square root of 6.

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How to find vertical and horizontal asymptotes of polynomials?
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Vertical asymptotes are when y tends to infinity, which in other words means you need a fraction where you're dividing by zero. So to find the vertical asymptotes, solve the denominator of the fraction and the solutions are the asymptotes since that is where it will be dividing by zero.

Horizontal asymptotes are the opposite and is where x tends to infinity -- This can be done by dividing through with the xs still there, since you are effectively just looking at multiples of infinity and infinity squared and so on. When you have an x to the power you can forget about all the other x's on the same half of the fraction, since infinity squared is infinitely bigger than infinity - yes it's abstract and a bit stupid I know.
If the degree of the polynomial on the denominator is larger than the numerator, there is an asymptote at y=0 (since the bottom becomes infinitely larger than the top). If it is lower than the numerator, you have to divide through by long division and you end up with an oblique asymptote i.e. an asymptote where the dotted line would follow a linear equation e.g. y=x

For example
f(x)= \frac{6x^2-3x+4}{2x^2-8}

The vertical asymptote can be found by solving 2x^2-8=0 i.e. x=+-2 so our vertical asymptotes are at 2 and -2.

The horizontal asymptote, think of all the x's as infinity. SInce there are x^2s, we dont need to worry about the numbers or the other x's
Therefore we are left with 6x^2/2x^2, which gives us 3. Therefore our horizontal asymptote would be at y=3.
7 0
3 years ago
4. Solve the equation.
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Step-by-step explanation:

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6 0
3 years ago
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4 years ago
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Answer:

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Step-by-step explanation:

Let's use distributive property to solve this equation.

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Answer: Triangle 2

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