All categories

3 years ago

HELP 12 pts!

Mathematics

2 answers:

OverLord2011 [107]3 years ago

5 0

Answer:

x = 2.4 cm

Step-by-step explanation:

HOPE I HELPED EVEN IF I ANSWERED MONTHS LATER :D

liq [111]3 years ago

4 0

The answer is “C” x=2.4. 1.2/.96=3.0/x. That equals 1.2x=2.88. 2.88/1.2=2.4

You might be interested in

If 12% of x is equal to 6% of y, then 18% of x will be equal to how much percent of y ?

Lelu [443]

Answer:

y=9%

Step-by-step explanation:

x=18%

y=9%

It is in a pattern like x=y+y

7 0

3 years ago

Read 2 more answers

Which statement is correct? A. -1.5 < -2 B. -2.5 < -3 C. -3.5 < -2.5

Pepsi [2]

Answer: C

Step-by-step explanation:

3 0

3 years ago

Write the equation of the line given the slope 2 and one point on the line (2, -4)

ELEN [110]

Point Slope Form:
$y - ( - 4) = 2(x - 2)$
Slope-intercept Form:
$y = 2x$

6 0

3 years ago

System with 3 variable. solve pleaseeee x+4y-5z=-7 3x+2y+3z=7 2x+y+5z=8

7nadin3 [17]

To solve this we need to form 2 equations with only 2 variables.
x+4y-5z=-7
2x+y+5z=8
3x+5y=1

15x+10y+15z=35
6x+3y+15z=24
9x+7y=11

Now we need to use these to find x and y
9x+15y=3
9x+7y=11
8y=-8
y=-1
Therefore:
x=2
From this we can find that:
2(2)-1+5z=8
Therefore z=1
The answer is x=2, y=-1, z=1

6 0

3 years ago

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