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4 years ago

Why does it take 3 copies of 1/6 to show the same amount as 1/2

2 answers:

5 0

Think of a pizza, cut in half. You have one half. If you cut each half into 3 pieces, you'll have a total of 6 pieces, so each piece is 1/6 of the pizza. But each half is made up of 3 1/6 pieces. So, three 1/6 pieces equals one 1/2 pizza.

Ann [662]4 years ago

3 0

Because 3/6 is the same as 1/2

You might be interested in

5 + (-6) = ? Math:adding with rational integers

REY [17]

5 + (-6) = -1
hope this helps!

8 0

4 years ago

Wich measurements is not precise 982g or 980g

gavmur [86]

Answer:

982 g

Step-by-step explanation:

Hopefully this helps you :)

8 0

3 years ago

Será a equação x² + 3x = x + 6 + x² do 2º grau?

goblinko [34]

x² + 3x = x + 6 + x²

-x² -x²

3x = x + 6

-x -x

2x = 6

$\frac{2x}{2}$ = $\frac{6}{2}$

x = 3

Answer: x = 3

3 0

3 years ago

Using power series, solve the LDE: (2x^2 + 1) y" + 2xy' - 4x² y = 0 --- - -- -

sattari [20]

We're looking for a solution of the form

$y=\displaystyle\sum_{n\ge0}a_nx^n$

with derivatives

$y'=\displaystyle\sum_{n\ge0}(n+1)a_{n+1}x^n$

$y''=\displaystyle\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n$

Substituting these into the ODE gives

$\displaystyle\sum_{n\ge0}\left(\bigg(2(n+2)(n+1)a_{n+2}-4a_n\bigg)x^{n+2}+2(n+1)a_{n+1}x^{n+1}+(n+2)(n+1)a_{n+2}x^n\right)=0$

Shifting indices to get each term in the summand to start at the same power of $x$ and pulling the first few terms of the resulting shifted series as needed gives

$2a_2+(2a_1+6a_3)x+\displaystyle\sum_{n\ge2}\bigg((n+2)(n+1)a_{n+2}+2n^2a_n-4a_{n-2}\bigg)x^n=0$

Then the coefficients in the series solution are given according to the recurrence

$\begin{cases}a_0=y(0)\\\\a_1=y'(0)\\\\a_2=0\\\\2a_1+6a_3=0\implies a_3=-\dfrac{a_1}3\\\\a_n=\dfrac{-2(n-2)^2a_{n-2}+4a_{n-4}}{n(n-1)}&\text{for }n\ge4\end{cases}$

Given the complexity of this recursive definition, it's unlikely that you'll be able to find an exact solution to this recurrence. (You're welcome to try. I've learned this the hard way on scratch paper.) So instead of trying to do that, you can compute the first few coefficients to find an approximate solution. I got, assuming initial values of $y(0)=y'(0)=1$ , a degree-8 approximation of

$y(x)\approx1+x-\dfrac{x^3}3+\dfrac{x^4}3+\dfrac{x^5}2-\dfrac{16x^6}{45}-\dfrac{79x^7}{125}+\dfrac{101x^8}{210}$

Attached are plots of the exact (blue) and series (orange) solutions with increasing degree (3, 4, 5, and 65) and the aforementioned initial values to demonstrate that the series solution converges to the exact one (over whichever interval the series converges, that is).

5 0

3 years ago

There are 10 blue marbles, 4 black marbles, 5 white marbles, and 6 red marbles in a box. If two marbles are drawn at random with

Nikitich [7]

Answer:

7/20

Step-by-step explanation:

10 blue marbles, 4 black marbles, 5 white marbles, and 6 red marbles = 25 marbles

25-10 = 15

There are 15 not blue marbles

P( not blue ) = not blue marbles / total = 15/25 = 3/5

keep the not blue marble

25-1 = 24 marbles left

15-1 = 14 not blue marbles

P( not blue ) = not blue marbles / total = 14/24 = 7/12

P( not blue, keep, not blue) = 3/5 * 7/12 = 21/60 = 7/20

6 0

4 years ago

Read 2 more answers