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Sonja [21]
3 years ago
6

Is

="latex-formula"> rational or irational. Why?
Mathematics
2 answers:
Natalka [10]3 years ago
7 0
The square root of 36 is a rational number. A rational number is a number that can be written as a fraction, a/b, where a and b are integers.
Akimi4 [234]3 years ago
3 0

\mathbf{Task. ~}\mathrm{Is} ~ \sqrt{36} ~ \mathrm{rational~or~irrational.~ Why?}

\mathbf{Rational~numbers~}\mathrm{are~numbers~that~can~be~represented~as~ordinary~ fractions.}

\mathrm{Examples \colon} ~ \dfrac{1}{2}; \ 0{,}5; \ -1\dfrac{2}{3} ; \ 0{,}(3).

\mathrm{All~integers~are~rational~numbers \colon} ~ 0; ~ 6; ~ 121; ~ 1 \ 000 \ 000; ~ {-5}; ~ {-41}

\mathbf{Irrational~numbers~}\mathrm{are~numbers~that~are~not~rational.}

\mathrm{Examples \colon} ~ \pi, ~ e, ~ \sqrt{3}, ~ \sqrt[3]{7}, ~ \log_{3}2

\mathrm{Since} ~ \sqrt{36} = 6 ~ \mathrm{then~it~is~a~rational~number.}

Answer \colon ~ \mathrm{rational~number}. ~ \blacktriangle

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3 years ago
Write the equation of the circle centered at ( 4 , − 5 ) with radius 18.
Free_Kalibri [48]

Answer:

(x-4)^2 + (y+5)^2 = 324

Step-by-step explanation:

The equation of a circle is given by the equation (x-h)^2 + (y-k)^2 = r^2, where (h,k) is the center point of the circle.

Therefore, since the center point of the circle and the radius is given, we can just plug the numbers into the formula:

(x-4)^2 + (y+5)^2 = 18^2

<u>(x-4)^2 + (y+5)^2 = 324</u>

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3 years ago
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3 0
2 years ago
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5 0
2 years ago
Simplify the expression state any excluded values<br> 2a^2-4a+2<br> ---------------<br> 3a^2-3
chubhunter [2.5K]

Answer:

The simplified form is \dfrac{2(x-1)}{3(x+1)}.

x =1 is the excluded value for the given expression.

Step-by-step explanation:

Given:

The expression given is:

\dfrac{2a^2-4a+2}{3a^2-3}

Let us simplify the numerator and denominator separately.

The numerator is given as 2a^2-4a+2

2 is a common factor in all the three terms. So, we factor it out. This gives,

=2(a^2-2a+1)

Now, a^2-2a+1=(a-1)(a-1)

Therefore, the numerator becomes 2(a-1)(a-1)

The denominator is given as: 3a^2-3

Factoring out 3, we get

3(a^2-1)

Now, a^2-1 is of the form a^2-b^2=(a-b)(a+b)

So, a^2-1=(a-1)(a+1)

Therefore, the denominator becomes 3(a-1)(a+1)

Now, the given expression is simplified to:

\frac{2a^2-4a+2}{3a^2-3}=\frac{2(x-1)(x-1)}{3(x-1)(x+1)}

There is (x-1) in the numerator and denominator. We can cancel them only if x\ne1 as for x=1, the given expression is undefined.

Now, cancelling the like terms considering x\ne1, we get:

\dfrac{2a^2-4a+2}{3a^2-3}=\dfrac{2(x-1)}{3(x+1)}

Therefore, the simplified form is \dfrac{2(x-1)}{3(x+1)}

The simplification is true only if  x\ne1. So, x =1 is the excluded value for the given expression.

8 0
3 years ago
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