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Sati [7]
3 years ago
11

How can you tell if a number is rational and irrational??

Mathematics
2 answers:
kompoz [17]3 years ago
4 0
All numbers<span> that are not </span>rational<span> are considered </span>irrational<span>. An </span>irrational number<span> can be written as a decimal, but not as a fraction. An </span>irrational number<span> has endless non-repeating digits to the right of the decimal point.</span>
evablogger [386]3 years ago
4 0
Easy..<span>All </span>numbers<span> that are not </span>rational<span> are considered </span>irrational<span>. An </span>irrational number<span> can be written as a decimal, but not as a fraction. An </span>irrational number<span> has endless non-repeating digits to the right of the decimal point.</span>
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What is the solution of the equation over the complex numbers
denis-greek [22]

<u>Answer:</u>

2i\sqrt{3}

-2i\sqrt{3}

<u>Step-by-step explanation:</u>

x^2 + 12

First set the equation to 0

x^2 + 12

x^2 + 12 = 0

Second, get the 12 on the right side of the equal sign by adding a -12 to each side

x^2 + 12 = 0

x^2 + 12 + (-12) = 0 -12

x^2 = -12

Square root both sides of the equal sign.

x^2 = -12

\sqrt{x^2} = \sqrt{-12}

Take the square root on left sides of the equal sign.

\sqrt{x^2} = \sqrt{-12}

x = \sqrt{-12}

Take the square root on the right side of the equal sign. Remember \sqrt{-1} = i

x = \sqrt{-12}

x = \sqrt{4}*\sqrt{-1}\sqrt{3}}

x = 2i\sqrt{3}

<u>AND</u>

x = -2i\sqrt{3}

8 0
3 years ago
PLS HELP ME ASAP FOR 42!! (MUST SHOW WORK!!) + LOTS OF POINTS!! *no calculator* *best if show work on picture*
Tpy6a [65]
It's 2.5 bc it goes up by 2.5 everytime
4 0
3 years ago
Directions: Find the indicated trigonometric ratio as a fraction in simplest form. 1. Sin L =
Rainbow [258]

Answer:

\sin L = 0.60

tan\ N = 1.33

\cos L = 0.80

\sin N = 0.80

Step-by-step explanation:

Given

See attachment

From the attachment, we have:

MN = 6

LN = 10

First, we need to calculate length LM,

Using Pythagoras theorem:

LN^2 = MN^2 + LM^2

10^2 = 6^2 + LM^2

100 = 36 + LM^2

Collect Like Terms

LM^2 = 100 - 36

LM^2 = 64

LM = 8

Solving (a): \sin L

\sin L = \frac{Opposite}{Hypotenuse}

\sin L = \frac{MN}{LN}

Substitute values for MN and LN

\sin L = \frac{6}{10}

\sin L = 0.60

Solving (b): tan\ N

tan\ N = \frac{Opposite}{Adjacent}

tan\ N = \frac{LM}{MN}

Substitute values for LM and MN

tan\ N = \frac{8}{6}

tan\ N = 1.33

Solving (c): \cos L

\cos L = \frac{Adjacent}{Hypotenuse}

\cos L = \frac{LM}{LN}

Substitute values for LN and LM

\cos L = \frac{8}{10}

\cos L = 0.80

Solving (d): \sin N

\sin N = \frac{Opposite}{Hypotenuse}

\sin N = \frac{LM}{LN}

Substitute values for LM and LN

\sin N = \frac{8}{10}

\sin N = 0.80

7 0
3 years ago
64% of 75 tiles is?
noname [10]
64\% \ of \ 75=64\% \times 75= \frac{64}{100} \times 75=\frac{16}{25} \times 75=16 \times 3=48

64% of 75 tiles is 48 tiles.
8 0
3 years ago
Read 2 more answers
Prove that 1/sin^2A -1/tan^2A= 1
Vika [28.1K]

Answer:

Step-by-step explanation:

LHS =\dfrac{1}{Sin^{2} \ A }-\dfrac{1}{Tan^{2} \ A }\\\\\\ = \dfrac{1}{sin^{2} \ A}- \dfrac{1}{\dfrac{Sin^{2} \ A}{Cos^{2} \ A}}\\\\\\= \dfrac{1}{sin^{2} \ A } - \dfrac{Cos^{2} \ A}{Sin^{2} \ A}\\\\\\= \dfrac{1-Cos^{2} \ A}{Sin^{2} \ A}\\\\\\= \dfrac{Sin^{2} \ A}{Sin^{2} \ A}\\\\\\= 1 = \ RHS

Hint: 1 - Cos² A = Sin² A

4 0
2 years ago
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