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aleksandrvk [35]
3 years ago
14

Rob pays $258 every 3 months for his karate lessons. At this rate, how much will Rob pay for 3 years of karate lessons?

Mathematics
2 answers:
otez555 [7]3 years ago
5 0
C is the answer
$3096
Afina-wow [57]3 years ago
5 0

If it is $258 for 3 months, there are 12 months in a year, multiply 258 by 4(to get to the total of 12 months) = $1032 per year, the question asks for 3 years, so you multiply 1032 by three to get $3096

The answer is C) $3096

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A scatter plot is shown:
gregori [183]

Answer:

B

Step-by-step explanation:

5 0
4 years ago
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All of the following expressions are equal except _____.
eimsori [14]
Well since this question is basically asking about Scientific Notation.

Scientific Notation-the way scientists easily handle large numbers or very small numbers.

for example:
0.00000000000000045 would be written as 4.5×10^-16
450000000000 would be written as 4.5×10^11
5 0
4 years ago
What is two numbers that add to 20 and the difference is 6
loris [4]

13 and 7.

Steps

x + x = 20

x - x = 6

2x = 26

x = 13

x = 20 - 13

x = 7

8 0
3 years ago
Read 2 more answers
13x+5y=90 what is x? what is y?
Leona [35]

Answer:

X=-5y/13+90/13, Y=-13x/5+18

Step-by-step explanation:

You need to solve the equation for X and Y.

Solving for X:

13x+5y=90

Subtract 5y: 13x=-5y+90

Divide by 13: x=-5y/13+90/13

--

For Y:

13x+5y=90

Subtract 13x: 5y=-13x+90

Divide by 5: y=-13x/5+90/5

(Simplifies to -13x/5+18)

3 0
3 years ago
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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