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NikAS [45]
3 years ago
8

I need help on .5 and .6

Mathematics
1 answer:
Dovator [93]3 years ago
8 0
5. 44 (forty four)
6. 58
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The figure below shows triangle NRM with r2 = m2 + n2:
alekssr [168]

Answer:

Option "SSS postulate" is correct.

Step-by-step explanation:

Given that: Triangle NRM has legs m and n, and r is the length of its longest side.

and r² = m² + n².

Now, Ben constructed a right triangle EFD with legs m and n.

and in statement 4, it is proved that f=r.

So, all the three sides of the triangles NMR and EFD are congruent.

So, the triangles are congruent bt the SSS Postulate.

⇒ option "SSS postulate" is correct.

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4 years ago
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What is the mid point between (0,0) and (3,4)
Maksim231197 [3]

Answer:

Midpoint of a line segment

(xM,yM) = (4, 5).

7 0
4 years ago
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Of the 24 students in gregs class, 3/8 ride the bus to school. how many students ride the bus?
Luda [366]
9 students ride the bus
5 0
4 years ago
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Make d the subject of this formula<br><br> H=d/3+2
Effectus [21]

Answer:

d= 3(h-2). hope you like it.

Step-by-step explanation:

4 0
3 years ago
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What's the intuition behind the equation 1+2+3+⋯=−1121+2+3+⋯=−112 ?
Aleks [24]
The sum clearly diverges. This is indisputable. The point of the claim above, that

1+2+3+\cdots=-\dfrac1{12}

is to demonstrate that a sum of infinitely many terms can be manipulated in a variety of ways to end up with a contradictory result. It's an artifact of trying to do computations with an infinite number of terms.

The mathematician Srinivasa Ramanujan famously demonstrated the above as follows: Suppose the series converges to some constant, call it C. Then

\begin{matrix}C&=&1&+2&+3&+4&+5&+6&+\cdots\\4C&=&&+4&&+8&&+12&+\cdots\\-3C&=&1&-2&+3&-4&+5&-6&+\cdots\end{matrix}

Now, recall the geometric power series

\displaystyle\sum_{n\ge0}x^n=1+x+x^2+x^3+\cdots=\dfrac1{1-x}

which holds for any |x|. It has derivative

\displaystyle\sum_{n\ge1}nx^{n-1}=1+2x+3x^2+4x^3+\cdots=\dfrac1{(1-x)^2}

Taking x=-1, we end up with

1+2(-1)+3(-1)^2+4(-1)^3+\cdots=1-2+3-4+\cdots=\dfrac14

and so

-3C=\dfrac14\implies C=-\dfrac1{12}

But as mentioned above, neither power series converges unless |x|. What Ramanujan did was to consider the sum 1-2+3-4+\cdots as a limit of the power series evaluated at x=-1:

\displaystyle-3C=\lim_{x\to-1^+}\sum_{n\ge1}nx^{n-1}=\lim_{x\to-1^+}\frac1{(1-x)^2}=\frac14

then arrived at the conclusion that C=-\dfrac1{12}.

But again, let's emphasize that this result is patently wrong, and only serves to demonstrate that one can't manipulate a sum of infinitely many terms like one would a sum of a finite number of terms.
4 0
3 years ago
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