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4 years ago

10,996 rounded to the neart thousand

Mathematics

2 answers:

vichka [17]4 years ago

8 0

Answer:

11000

Step-by-step explanation:

Find the number in the thousand place

and look one place to the right for the rounding digit

. Round up if this number is greater than or equal to

and round down if it is less than

Sholpan [36]4 years ago

3 0

<h3>Rounding it to the nearest thousand:</h3>

11,000

<h2>Answer:</h2>

11,000

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Water leaks from a crack in a vase at a rate of 0.5 cubic inch per minute. How long does it take for 20% of the water to leak fr

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Find total volume to find what amount is 20%
hack, is put the 20% in alreadyy to find 20% volume
20%=1/5

so
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4.8=d
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4 years ago

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3 years ago

Find the general solution of the nonhomogeneous differential equation x^2y''-2y=3(x^2) -1, (x>0).

Ira Lisetskai [31]

Answer:

G.S= $C_1\frac{1}{x}+C_2x^2+x^2logx+\frac{1}{2}$

Step-by-step explanation:

We are given that non-homogeneous differential equation

$x^2y''-2y=3(x^2)-1$

It is Cauchy Euler equation

Substitute x=e^t x>0

Auxillary equation

$D'(D'-1)-2=0$

$D'^2-D'-2=0$

$(D'-2)(D'+1)=0$

$D'-2=0 \implies D'=2$

$D'+1=0\implies D'=-1$

Complementary solution

$y=C_1e^{-t}+C_2e^{2t}$

$y=C_1\frac{1}{x}+C_2x^2$

Particular solution

$y_p=\frac{3e^{2t}}{D'^2-D'-2}-\frac{e^{0t}}{D'^2-D'-2}$

$y_p=te^{2t}+\frac{1}{2}=x^2logx+\frac{1}{2}$

G.S= $C_1\frac{1}{x}+C_2x^2+x^2logx+\frac{1}{2}$

Hence, general solution G.S= $C_1\frac{1}{x}+C_2x^2+x^2logx+\frac{1}{2}$

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4 years ago

A group issimpleif it has no nontrivial proper normal subgroups. LetGbe a simple group of order168. How many elements of order 7

stepladder [879]

Answer:

6*8=48 groups with elements of order 7

Step-by-step explanation:

For this case the first step is discompose the number 168 in factors like this:

$168 = 8*3*7= 2^3 *3*7$

And for this case we can use the Sylow theorems, given by:

Let G a group of order $p^{\alpha} m$ where p is a prime number, with $m\leq 1$ and p not divide m then:

1) $Syl (G) \neq \emptyset$

2) All sylow p subgroups are conjugate in G

3) Any p subgroup of G is contained in a Sylow p subgroup

4) n(G) =1 mod p

Using these theorems we can see that 7 = 1 (mod7)

By the theorem we can't have on one Sylow 7 subgroup so then we need to have 8 of them.

Every each 2 subgroups intersect in a subgroup with a order that divides 7. And analyzing the intersection we can see that we can have 6 of these subgroups.

So then based on the information we can have 6*8=48 groups with elements of order 7 in G of size 168

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3 years ago