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3 years ago

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

2 answers:

4 0

All numbers that are not rational are considered irrational. An irrational number can be written as a decimal, but not as a fraction. An irrational number has endless non-repeating digits to the right of the decimal point.

evablogger [386]3 years ago

4 0

Easy..All numbers that are not rational are considered irrational. An irrational number can be written as a decimal, but not as a fraction. An irrational number has endless non-repeating digits to the right of the decimal point.

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What is the solution of the equation over the complex numbers

denis-greek [22]

Answer:

$2i\sqrt{3}$

$-2i\sqrt{3}$

Step-by-step explanation:

$x^2 + 12$

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 $\sqrt{-1} = i$

$x = \sqrt{-12}$

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

$x = 2i\sqrt{3}$

AND

$x = -2i\sqrt{3}$

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

3 years ago