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Degger [83]
3 years ago
13

Help I don't know any of this 

Mathematics
2 answers:
leonid [27]3 years ago
8 0
3.1415... is not a rational number because it is continuous.

Hope I helped :)
andrezito [222]3 years ago
4 0
Ratinonal numbers can repeat as long we can see the repeating part with a line over it (called a vinculum)
rational numbers can be written in form a/b where b is not equal zero

1.33333333333333 is repeating, it is 4/3

3.1415... is obviously an attempt to represent pi or π, the most famous irrational number

5.25 is easy, 5.25/1 or 525/100

0.6666666666666=2/3

the answer is 3.141592...
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5 0
3 years ago
Please help me <br> Show your work <br> 10 points
Svet_ta [14]
<h2>Answer</h2>

After the dilation \frac{5}{3} around the center of dilation (2, -2), our triangle will have coordinates:

R'=(2,3)

S'=(2,-2)

T'=(-3,-2)

<h2>Explanation</h2>

First, we are going to translate the center of dilation to the origin. Since the center of dilation is (2, -2) we need to move two units to the left (-2) and two units up (2) to get to the origin. Therefore, our first partial rule will be:

(x,y)→(x-2, y+2)

Next, we are going to perform our dilation, so we are going to multiply our resulting point by the dilation factor \frac{5}{3}. Therefore our second partial rule will be:

(x,y)→\frac{5}{3} (x-2,y+2)

(x,y)→(\frac{5}{3} x-\frac{10}{3} ,\frac{5}{3} y+\frac{10}{3} )

Now, the only thing left to create our actual rule is going back from the origin to the original center of dilation, so we need to move two units to the right (2) and two units down (-2)

(x,y)→(\frac{5}{3} x-\frac{10}{3}+2,\frac{5}{3} y+\frac{10}{3}-2)

(x,y)→(\frac{5}{3} x-\frac{4}{3} ,\frac{5}{3}y+ \frac{4}{3})

Now that we have our rule, we just need to apply it to each point of our triangle to perform the required dilation:

R=(2,1)

R'=(\frac{5}{3} x-\frac{4}{3} ,\frac{5}{3}y+ \frac{4}{3})

R'=(\frac{5}{3} (2)-\frac{4}{3} ,\frac{5}{3}(1)+ \frac{4}{3})

R'=(\frac{10}{3} -\frac{4}{3} ,\frac{5}{3}+ \frac{4}{3})

R'=(2,3)

S=(2,-2)

S'=(\frac{5}{3} (2)-\frac{4}{3} ,\frac{5}{3}(-2)+ \frac{4}{3})

S'=(\frac{10}{3} -\frac{4}{3} ,-\frac{10}{3}+ \frac{4}{3})

S'=(2,-2)

T=(-1,-2)

T'=(\frac{5}{3} (-1)-\frac{4}{3} ,\frac{5}{3}(-2)+ \frac{4}{3})

T'=(-\frac{5}{3} -\frac{4}{3} ,-\frac{10}{3}+ \frac{4}{3})

T'=(-3,-2)

Now we can finally draw our triangle:

8 0
3 years ago
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