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2 years ago

What are some equivalent expressions of 0.003 x 0.00009

Mathematics

1 answer:

QveST [7]2 years ago

6 0

Answer:

3x0.9, 30x9, 3/1000 times 9/10000

Step-by-step explanation:

Multiply each factor by multiples of ten.

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What is the slope of the line passing through the points (2, 5) and (0, –4)?

iragen [17]

In order to find the slope of line that passes through 2 points, use the equation slope=rise/run
rise = 5-(-4) =9
run = 2- 0 =2
slope = rise/run = 9/2

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2 years ago

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Use the complement to find the probability. Enter your answer in simplified fraction form.

Ket [755]

2/3 or 66.666666...% is the probability of it not landing on blue

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2 years ago

Plz answer letter b and number 5.6

blsea [12.9K]

Answer:

b. Is because -3 because the the negative gets turned into a positive.

5. They are almost the same because subtraction change the negative to a positive and the -5+2 is the aftermath of what happend.

6. Its because the subtract a negative it cancels out and makes the number a positive.

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2 years ago

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The perimeter of the rectangle below is 130 units. Find the value of z

aev [14]

Answer:

9 =z

Step-by-step explanation:

The perimeter of a rectangle is

P = 2(l+w) where l is the length and w is the width

P = 2(3z+2+ 4z)

Combine like terms

P = 2(7z+2)

Substitute in the perimeter

130 = 2( 7z+2)

Divide each side by 2

130/2 = 2/2 ( 7z+2)

65 = 7z+2

Subtract 2 from each side

65-2 = 7z+2-2

63 = 7z

Divide by 7

63/7 = 7z/7

9 =z

8 0

2 years ago

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For the differential equation 3x^2y''+2xy'+x^2y=0 show that the point x = 0 is a regular singular point (either by using the lim

Svetlanka [38]

Given an ODE of the form

$y''(x)+p(x)y'(x)+q(x)y(x)=f(x)$

a regular singular point $x=c$ is one such that $p(x)$ or $q(x)$ diverge as $x\to c$ , but the limits of $(x-c)p(x)$ and $(x-c)^2q(x)$ as $x\to c$ exist.

We have for $x\neq0$ ,

$3x^2y''+2xy'+x^2y=0\implies y''+\dfrac2{3x}y'+\dfrac13y=0$

and as $x\to0$ , we have $x\cdot\dfrac2{3x}\to\dfrac23$ and $x^2\cdot\dfrac13\to0$ , so indeed $x=0$ is a regular singular point.

We then look for a series solution about the regular singular point $x=0$ of the form

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

Substituting into the ODE gives

$\displaystyle3x^2\sum_{n\ge0}a_n(n+k)(n+k-1)x^{n+k-2}+2x\sum_{n\ge0}a_n(n+k)x^{n+k-1}+x^2\sum_{n\ge0}a_nx^{n+k}=0$

$\displaystyle3\sum_{n\ge2}a_n(n+k)(n+k-1)x^{n+k}+3a_1k(k+1)x^{k+1}+3a_0k(k-1)x^k$
$\displaystyle+2\sum_{n\ge2}a_n(n+k)x^{n+k}+2a_1(k+1)x^{k+1}+2a_0kx^k$
$\displaystyle+\sum_{n\ge2}a_{n-2}x^{n+k}=0$

From this we find the indicial equation to be

$(3(k-1)+2)ka_0=0\implies k=0,\,k=\dfrac13$

Taking $k=\dfrac13$ , and in the $x^{k+1}$ term above we find $a_1=0$ . So we have

$\begin{cases}a_0=1\\a_1=0\\\\a_n=-\dfrac{a_{n-2}}{n(3n+1)}&\text{for }n\ge2\end{cases}$

Since $a_1=0$ , all coefficients with an odd index will also vanish.

So the first three terms of the series expansion of this solution are

$\displaystyle\sum_{n\ge0}a_nx^{n+1/3}=a_0x^{1/3}+a_2x^{7/3}+a_4x^{13/3}$

with $a_0=1$ , $a_2=-\dfrac1{14}$ , and $a_4=\dfrac1{728}$ .

6 0

3 years ago