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densk [106]
3 years ago
12

Which irrational number can be added to π to get a sun that is rational?

Mathematics
1 answer:
MatroZZZ [7]3 years ago
3 0
The answer is C.

Adding any positive value to \pi won't really make \pi into a rational number, so the way you have to do it add the negative self, -\pi to make it into 0. Any number plus its negative value would result in 0, which is a rational number.
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Please explain your answer as well. Thx!!!
dimulka [17.4K]

Answer:

e^2

Step-by-step explanation:

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Chapter Test Find the distance between (2, -1) and (3, 4). The distance is units​
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Answer:

657483920

Step-by-step explanation:

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What is the missing reason in Step 8? 
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The missing reason is you didn't put the actual question on here.

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What is 9 2/3 - 2 8/10
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Find the other two vertices of a square with one vertex (0, 0) and another vertex (4, 2). Can you find another answer?
Margarita [4]

It is given that two vertices of square are (0,0) and (4,2).

Now the problem is that you haven't given that whether these two vertices are adjacent vertices or opposite vertices of the square.

1. By Supposing that these two are adjacent vertices of Square

The third vertex will be at (-4,2) which lies in third quadrant.

Suppose the coordinate of fourth vertex be (x,y).

Mid point of line joining (4,2) and (-4,2) is{ [4+(-4)]/2,(2+2)/2} is (0,2).

Mid point of line joining (x,y) and (0,0) is (x/2,y/2).

Since diagonals of square bisect each other,

∵ x/2=0

⇒x=0

and

y/2=2

⇒y=4

So, The Coordinate of  fourth vertex is (0,4).

Now coming back to second condition if these are two opposite vertex of Square.

Let the third coordinate be (a,b).

Length of diagonal=\sqrt{(4-0)^2+(2-0)^2}=\sqrt20=2\sqrt5

Now,let side of Square be A.

Then length of Diagonal of square =√2 A

⇒√2 A=2√5

⇒A =√10

As third vertex is (a,b).

Using distance formula

a² + b²=10  -------------(1)

(a-4)²+ (b-2)²=10  --------------(2)

Solving expression (1) and (2), we get

⇒a²+ b²=(a-4)² +(b-2)²

⇒2a + b =5

⇒b=5-2a

Putting the value of b in (1),we get

⇒a² +(5-2a)²=10

⇒a²+25+4a²-20a =10

⇒5a²-20a+15=0

⇒a² - 4a + 3=0

Splitting the middle term,we get

⇒(a-3)(a-1)=0

⇒a=3  ∧  a=1

we get b=5-2×1=3 and b=5-2×3=5-6=-1

So,the other vertex are (1,3) and(3,-1).





3 0
3 years ago
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