Is 0.2314 rational or irrational

Mathematics

1 answer:

professor190 [17]4 years ago

6 0

Irrational: a number that is nonterminating and nonrepeating decimal

Rational: a number that can be a fraction that has the numerator and denominator as integers, is terminating, can be a repeating decimal.

So its irrational

Hope that helps :)

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PLEASE HELP ASAP! ALGEBRA!<br><br> (0,5), m=1/3 in point slope form

n200080 [17]

Answer:

y - 5 = 1/3(x - 0)

Step-by-step explanation:

the formula for point-slope form is y2 - y1 = m(x2 - x1), where m is the slope, y1 is the y coordinate and x1 is the x coordinate

so just plug in the values into the equation:

y - 5 = 1/3(x - 0)

if you want it in slope-intercept form, the formula is: y = mx+b, where mx is the slope and b is the y intercept. it would look like the following:

y = 1/3x + 5

5 0

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Sketch the graph: y=1/3x+2

Cerrena [4.2K]

The slope of the line is 1/3and the y-intercept is at the (0, 2) as shown in the graph below.

<h3>Graph of a linear function</h3>

A linear function is a function that has a leading degree of 1. The standard equation for a linear function is expressed as:

y = mx + b

where

m is the slope which is equal to the rate of change of y coordinate to x-coordinates.

b is the y-intercept

Given the following equation

y= 1/3x + 2

The slope of the line is 1/3 and the y-intercept is at the(0,2) as shown in the graph below.

Learn more on equation of a line here: brainly.com/question/13763238

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6 0

2 years ago

Find the six trigonometric function values for angle ∅ where its adjacent side is -9 and its hypotenuse is 41. (Theta is located

arsen [322]

Check the picture below.

$\bf \textit{using the pythagorean theorem}
\\\\
c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\sqrt{41^2-(-9)^2}=b\implies \sqrt{1681-81}=b\\\\\\ \sqrt{1600}=b\implies 40=b\\\\
-------------------------------$

$\bf sin(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{hypotenuse}{41}}\qquad~~ cos(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{hypotenuse}{41}}\qquad~~ tan(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{adjacent}{-9}}
\\\\\\
csc(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{opposite}{40}}\qquad ~~sec(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{adjacent}{-9}}\qquad ~~cot(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{opposite}{40}}$