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2 years ago

Which digit in 1,377,207 has a value that is 1/10 the value of the 7 in the ten thousand place?

1 answer:

5 0

Considering the values of the digits in the number, we have that digit 2 has a value that is 1/10 the value of the 7 in the ten thousand place.

<h3>What are the values of the digits in the number 1,377,207?</h3>

We start counting the values from the final number, from 1 and multiplying by 10, hence:

Digit 7 has a value of 1.

Digit 0 has a value of 10.

Digit 2 has a value of 100.

Digit 7 has a value of 1,000.

Digit 7 has a value of 10,000. -> Ten thousand place.

Digit 3 has a value of 100,000.

Digit 2 has a value of 1,000,000.

One tenth of the ten thousand place is a value of 100, hence digit 2 has a value that is 1/10 the value of the 7 in the ten thousand place.

More can be learned about the values of the digits in a number at brainly.com/question/2041524

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3 years ago

Solve for y where y(2)=2 and y'(2)=0 by representing y as a power series centered at x=a

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I'll assume the ODE is actually

$y''+(x-2)y'+y=0$

Look for a series solution centered at $x=2$ , with

$y=\displaystyle\sum_{n\ge0}c_n(x-2)^n$

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

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

with $c_0=y(2)=2$ and $c_1=y'(2)=0$ .

Substituting the series into the ODE gives

$\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge0}(n+1)c_{n+1}(x-2)^{n+1}+\sum_{n\ge0}c_n(x-2)^n=0$

$\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge1}nc_n(x-2)^n+\sum_{n\ge0}c_n(x-2)^n=0$

$\displaystyle2c_2+c_0+\sum_{n\ge1}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge1}nc_n(x-2)^n+\sum_{n\ge1}c_n(x-2)^n=0$

$\displaystyle2c_2+c_0+\sum_{n\ge1}\bigg((n+2)(n+1)c_{n+2}+(n+1)c_n\bigg)(x-2)^n=0$

$\implies\begin{cases}c_0=2\\c_1=0\\(n+2)c_{n+2}+c_n=0&\text{for }n>0\end{cases}$

If $n=2k$ for integers $k\ge0$ , then

$k=0\implies n=0\implies c_0=c_0$

$k=1\implies n=2\implies c_2=-\dfrac{c_0}2=(-1)^1\dfrac{c_0}{2^1(1)}$

$k=2\implies n=4\implies c_4=-\dfrac{c_2}4=(-1)^2\dfrac{c_0}{2^2(2\cdot1)}$

$k=3\implies n=6\implies c_6=-\dfrac{c_4}6=(-1)^3\dfrac{c_0}{2^3(3\cdot2\cdot1)}$

and so on, with

$c_{2k}=(-1)^k\dfrac{c_0}{2^kk!}$

If $n=2k+1$ , we have $c_{2k+1}=0$ for all $k\ge0$ because $c_1=0$ causes every odd-indexed coefficient to vanish.

So we have

$y(x)=\displaystyle\sum_{k\ge0}c_{2k}(x-2)^{2k}=\sum_{k\ge0}(-1)^k\frac{(x-2)^{2k}}{2^{k-1}k!}$

Recall that

$e^x=\displaystyle\sum_{n\ge0}\frac{x^k}{k!}$

The solution we found can then be written as

$y(x)=\displaystyle2\sum_{k\ge0}\frac1{k!}\left(-\frac{(x-2)^2}2\right)^k$

$\implies\boxed{y(x)=2e^{-(x-2)^2/2}}$

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4 years ago

What is the value of...?

ELEN [110]

Step 1
calculate the value for i=1
=(4*(1/2)<span> ^(1-1))---> 4

step 2
</span>calculate the value for i=2
=(4*(1/2) ^(2-1))---> 2

step 3
calculate the value for i=
=(4*(1/2) ^(3-1))---> 4*(1/4)---->1

step 4
find the sum
(4+2+1)=7

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3 years ago