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dsp73
2 years ago
15

Is √13 + 4 rational or irrational?

Mathematics
2 answers:
ycow [4]2 years ago
6 0

Hi student, let me help you out! :)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

We are asked to find out if \pmb{\sqrt{13} +4} is rational or irrational.

\triangle~\fbox{\bf{KEY:}}

  • Irrational numbers <u>cannot </u>be expressed as fractions (in \pmb{\cfrac{p}{q}} form).

So, can we write \pmb{\sqrt{13} +4} as a fraction? We can write 4 as a fraction, but we cannot write \pmb{\sqrt{13}} as a fraction, because this number has infinitely many digits after the decimal point, and they <u><em>never</em></u> repeat; there's no pattern at all!

And even if we add 4, which is a rational number, to \pmb{\sqrt{13}}, we'll still get an irrational number.

Thus, \pmb{\sqrt{13}+4} is irrational.

Hope it helps you out! :D

Ask in comments if any queries arise.

#StudyWithBrainly

~Just a smiley person helping fellow students out :)

LiRa [457]2 years ago
5 0

Answer:

irrational

Step-by-step explanation:

what are rational numbers ?

every number that can be expressed as a/b, with a and b being integer numbers, and b different to 0.

the square root of any number that is not a squared rational number is then an irrational number.

I am not sure, if the expression here is

sqrt(13) + 4

or

sqrt(13 + 4), which would then be sqrt(17).

but in both cases we are creating the square root of a not- squared rational number.

neither 13 nor 17 are the result of a multiplication of a rational number with itself.

in fact, 13 and 17 are prime numbers with no other factors than either 1 or themselves.

so, both, sqrt(13) and sqrt(17) are irrational numbers.

adding or subtracting rational numbers to irrational numbers result again in irrational numbers.

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Vikentia [17]
We have a base 2 with an exponent power raised to -3
so in order to have a positive exponent factor, we take reciprocal of the number1/2^3.
now that the base has a positive exponent number, we simply multiply it the times on the exponent number, so: 1/2*2*2 = 1/8

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5 0
3 years ago
HELp meh with this question it very hard
Dovator [93]

Answer:

  • AB = 7 cm

Step-by-step explanation:

<u>Use the law of cosines to find the side AB:</u>

  • AB = \sqrt{(x + 3)^2+x^2-2x(x+3)cos (60)} =
  • \sqrt{x^2+6x+9+x^2-x^2-3x}  = \sqrt{x^2+3x+9}

<u>Use the Heron's area formula next:</u>

  • A = \sqrt{s(s - a)(s-b)(s-c)}, where s- semi perimeter
  • s = 1/2[x + x + 3 + \sqrt{x^2+3x+9}) = 1/2 (2x + 3 + \sqrt{x^2+3x+9})
  • s - a = 1/2 (2x + 3 + \sqrt{x^2+3x+9} - 2x - 6) = 1/2 (\sqrt{x^2+3x+9 } - 3)
  • s - b = 1/2 (2x + 3 + \sqrt{x^2+3x+9} - 2x) = 1/2 (\sqrt{x^2+3x+9} + 3)
  • s - c = 1/2 (2x + 3 + \sqrt{x^2+3x+9} - 2\sqrt{x^2+3x+9}) = 1/2 (2x + 3 - \sqrt{x^2+3x+9})

<u>Now</u>

  • (s - a)(s - b) = 1/4 [(x²+3x+9) - 9] = 1/4 (x² + 3x)
  • s(s - c) = 1/4 [(2x + 3)² - (x² + 3x + 9)] = 1/4 (3x²+ 9x) = 3/4(x² + 3x)

<u>Next</u>

  • A² = 3/16(x² + 3x)(x² + 3x)
  • 300 = 3/16(x² + 3x)²
  • 1600 = (x² + 3x)²
  • x² + 3x = 40

<u>Substitute this into the first equation:</u>

  • AB = \sqrt{40 + 9} = 7 cm

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