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MAXImum [283]
3 years ago
15

Is -3/2 a rational or irrational number?

Mathematics
1 answer:
lianna [129]3 years ago
7 0

Answer: rational

Step-by-step explanation:

simplfy first and then it is  an intger and deimal so it makes it a rational when u simplfy u get 1/5

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The mistake is in step 3:

I: −2m =−12 − 2n
II: 2m = 8 + 8n

the student set both right sides of the equations equal (−12 − 2n = 8 + 8n), but the left sides aren't equal: first equation has -2m, second has +2m so the signs of one of the equations must be inverted first
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3 years ago
Raphael graphed the functions g (x)=x+2 and f (x)= x-1. how many units below the y-intercept of g (x) is the y-intercept of f (x
Luda [366]
<span>In our equations, you can use the generic form of y = mx + b to determine the y-intercept for the function, with b equal to the y-intercept. For g(x), b =2 and for f(x), b=-1. These values are the y-intercepts for the functions. Based on this, the y-intercept of f(x) is 3 units below the y-intercept of g(x). We know this because we can subtract the b value from f(x) from g(x) to get the difference. Difference = 2 - (-1) = 3.</span>
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3 years ago
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Please help! (question in attachment)
igor_vitrenko [27]
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5 0
2 years ago
In a survey of 3,000 people who owned a certain type of​ car, 1,050 said they would buy that type of car again. What percent of
Ray Of Light [21]

Answer:

35 percent

Step-by-step explanation:

1050 / 3000 as a percent

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300 / 3 = 100

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3 0
2 years ago
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Is sin theta=5/6, what are the values of cos theta and tan theta?
mina [271]

let's bear in mind that sin(θ) in this case is positive, that happens only in the I and II Quadrants, where the cosine/adjacent are positive and negative respectively.

\bf sin(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{hypotenuse}{6}}\qquad \impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{6^2-5^2}=a\implies \pm\sqrt{36-25}\implies \pm \sqrt{11}=a \\\\[-0.35em] ~\dotfill

\bf cos(\theta )=\cfrac{\stackrel{adjacent}{\pm\sqrt{11}}}{\stackrel{hypotenuse}{6}} \\\\\\ tan(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{adjacent}{\pm\sqrt{11}}}\implies \stackrel{\textit{and rationalizing the denominator}~\hfill }{tan(\theta )=\pm\cfrac{5}{\sqrt{11}}\cdot \cfrac{\sqrt{11}}{\sqrt{11}}\implies tan(\theta )=\pm\cfrac{5\sqrt{11}}{11}}

5 0
2 years ago
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