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miskamm [114]
3 years ago
5

HELP ME RNRNRN I WIll give brainlist

Mathematics
1 answer:
Sliva [168]3 years ago
3 0

Answer:

5^2=25, so a=5

Step-by-step explanation:

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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
to design the interior surface of a huge stainless-steel tank, you revolve the curve y=x^2 on the interval 0x4 about the y-axis.
Oksana_A [137]

Answer:

W = 21.44*10⁶J

Step-by-step explanation:

Given

y = x²    (0 < x < 4)

γ = 10000 N/m³

W = ?

If we apply

y = 4² = 16 = h

y =  x²  ⇒  x = √y

then

V = π*(√y)²*dy = π*y*dy

W = (γ*V) *(16-y) = γ*π*y*dy*(16-y)

⇒  W = γ*π*∫y*(16-y)dy = γ*π*(8y²-(y³/3))

finally we obtain  (0 < y < 16)

W = γ*π*(8y²-(y³/3)) = 10000*π*(8*16²-(16³/3)) = 21.44*10⁶J

7 0
3 years ago
For what value(s) of k will the relation not be a function?
Roman55 [17]

Answer:

b=k3+24k2−3700k

r=(k3−5k2+3k−5)((−k)(4))

87(|k+1|)+2088

Step-by-step explanation:

4 0
3 years ago
Which fraction is NOT equivalent to
Alenkinab [10]

1/3

7/12

9/12

1/4

and more

3 0
3 years ago
4 1/3 - 1 2/3 how to solve this please
8_murik_8 [283]

Answer:

2\frac{2}{3}

Step-by-step explanation:

1) Convert 4\frac{1}{3} to improper fraction. Use this rule: a\frac{b}{c} =\frac{ac+b}{c}.

\frac{4\times3+1}{3} -1\frac{2}{3}

2) Simplify 4 * 3 to 12.

\frac{12+1}{3}

3) Simplify 12 + 1 to 13.

\frac{13}{3} -1\frac{2}{3}

4) Convert 1\frac{2}{3}   to improper fraction. Use this rule: a\frac{b}{c} =\frac{ac+b}{c}.

\frac{13}{3} -\frac{1\times3+2}{3}

5) Simplify 1 * 3 to 3.

\frac{13}{3} -\frac{3+2}{3}

6) Simplify 3 + 2 to 5.

\frac{13}{3} -\frac{5}{3}

7) Join the denominators.

\frac{13-5}{3}

8) Simplify.

\frac{8}{3}

9) Convert to mixed fraction.

2\frac{2}{3}

(Decimal Form: 2.666667)

Thank you,
Eddie

3 0
2 years ago
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