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2 years ago

Simplify the equation

{245c^3}" align="absmiddle" class="latex-formula">

Mathematics

2 answers:

Simora [160]2 years ago

5 0

Answer:

mhm-

Step-by-step explanmhm-

Flauer [41]2 years ago

5 0

√ 245c^3 Simplified, should look like this.

Answer: 7c √5c

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Use the method of undetermined coefficients to solve the given nonhomogeneous system. x' = −1 5 −1 1 x + sin(t) −2 cos(t)

AlekseyPX

It looks like the system is

$x' = \begin{bmatrix} -1 & 5 \\ -1 & 1 \end{bmatrix} x + \begin{bmatrix} \sin(t) \\ -2 \cos(t) \end{bmatrix}$

Compute the eigenvalues of the coefficient matrix.

$\begin{vmatrix} -1 - \lambda & 5 \\ -1 & 1 - \lambda \end{vmatrix} = \lambda^2 + 4 = 0 \implies \lambda = \pm2i$

For $\lambda = 2i$ , the corresponding eigenvector is $\eta=\begin{bmatrix}\eta_1&\eta_2\end{bmatrix}^\top$ such that

$\begin{bmatrix} -1 - 2i & 5 \\ -1 & 1 - 2i \end{bmatrix} \begin{bmatrix} \eta_1 \\ \eta_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

Notice that the first row is 1 + 2i times the second row, so

$(1+2i) \eta_1 - 5\eta_2 = 0$

Let $\eta_1 = 1-2i$ ; then $\eta_2=1$ , so that

$\begin{bmatrix} -1 & 5 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 - 2i \\ 1 \end{bmatrix} = 2i \begin{bmatrix} 1 - 2i \\ 1 \end{bmatrix}$

The eigenvector corresponding to $\lambda=-2i$ is the complex conjugate of $\eta$ .

So, the characteristic solution to the homogeneous system is

$x = C_1 e^{2it} \begin{bmatrix} 1 - 2i \\ 1 \end{bmatrix} + C_2 e^{-2it} \begin{bmatrix} 1 + 2i \\ 1 \end{bmatrix}$

The characteristic solution contains $\cos(2t)$ and $\sin(2t)$ , both of which are linearly independent to $\cos(t)$ and $\sin(t)$ . So for the nonhomogeneous part, we consider the ansatz particular solution

$x = \cos(t) \begin{bmatrix} a \\ b \end{bmatrix} + \sin(t) \begin{bmatrix} c \\ d \end{bmatrix}$

Differentiating this and substituting into the ODE system gives

$-\sin(t) \begin{bmatrix} a \\ b \end{bmatrix} + \cos(t) \begin{bmatrix} c \\ d \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ -1 & 1 \end{bmatrix} \left(\cos(t) \begin{bmatrix} a \\ b \end{bmatrix} + \sin(t) \begin{bmatrix} c \\ d \end{bmatrix}\right) + \begin{bmatrix} \sin(t) \\ -2 \cos(t) \end{bmatrix}$

$\implies \begin{cases}a - 5c + d = 1 \\ b - c + d = 0 \\ 5a - b + c = 0 \\ a - b + d = -2 \end{cases} \implies a=\dfrac{11}{41}, b=\dfrac{38}{41}, c=-\dfrac{17}{41}, d=-\dfrac{55}{41}$

Then the general solution to the system is

$x = C_1 e^{2it} \begin{bmatrix} 1 - 2i \\ 1 \end{bmatrix} + C_2 e^{-2it} \begin{bmatrix} 1 + 2i \\ 1 \end{bmatrix} + \dfrac1{41} \cos(t) \begin{bmatrix} 11 \\ 38 \end{bmatrix} - \dfrac1{41} \sin(t) \begin{bmatrix} 17 \\ 55 \end{bmatrix}$

7 0

1 year ago

I need help if you get it right I will mark brainliest pls

mel-nik [20]

The answer is.......5/3

7 0

2 years ago

Can someone help me with this problem?

sineoko [7]

13/3 would be the answer :3

7 0

3 years ago

Read 2 more answers

The ruthless queen has asked the royal physician to come

ANTONII [103]

The amount of ounces of the 7.9% substance is 2 ounces

<h3>How to determine the how many ounces of the 7.9% he used?</h3>

We know that amount m = CV where

c = concentration and
V = volume

Let

m = amount of 3.7 % ,
m' = amount of 7.9 % and
M = amount of 6.5 %

We have that m + m' = M

cv + c'v' = CV where

c = 3.7 % conentration = 0.037
v = volume of 3.7 % = 1 ounce ,
c = 7.9% concentration = 0.079,
v' = volume of 7.9%,
C = 6.5 % concentration = 0.065 and
V = volume of 6.5% = v + v'

Since cv + c'v' = CV

Substituting the values of the variables into the equation, we have

3.7% × 1/3.7% + 7.9 %v' = 6.5%(v + v')

3.7% + 7.9 %v' = 6.5%(1 + v')

3.7% + 7.9 %v' = 6.5% + 6.5%v'

7.9%v' - 6.5%v' = 6.5% - 3.7%

1.4%v' = 2.8%

v' = 2.8%/1.4%

v' = 2 ounces

So, the amount of ounces of the 7.9% substance is 2 ounces

Learn more about amount of ounces here:

brainly.com/question/2799361

#SPJ1

6 0

1 year ago

While waiting for a video game to download, you notice that 30 percent of 32000 kilobytes have been downloaded so far. How many

Gnesinka [82]

Answer

Find out the how many kilobytes have been downloaded so far .

To prove

As given

While waiting for a video game to download.

notice that 30 percent of 32000 kilobytes have been downloaded so far .

30% is written in the decimal form.

$= \frac{30}{100}$

= 0.30

kilobytes uses in downloaded = 0.30 × 32000

= 9600 kilobytes

Therefore 9600 kilobytes have been downloaded so far .

6 0

3 years ago

Read 2 more answers