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mrs_skeptik [129]

3 years ago

9

Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.

1 answer:

irakobra [83]3 years ago

6 0

Answer:

a-9=20

a=20+9

a=29

b-9>20

b>29

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G(n)= 2n-3 f(n)= 2n + 1 Find (g•f)(n)

Vitek1552 [10]

Answer:

Step-by-step explanation:

(2n -3)(2n+1) = 4n^2 + 2n - 6n - 3 = 4n^2 - 4n -3

6 0

4 years ago

PLS HELP ASAP if you could explain how to do this as well im really slow when it comes to math :/

morpeh [17]

Answer:f(x)= 3*1/k

8=3*1/k

K=3/8

When x=2

f(x)=2*3/8

f(x)=3/4

Step-by-step explanation:

5 0

3 years ago

Solve the system.<br> x+3y= 22<br> 2x -y=2<br> (7,5)<br> (2,2)<br> (4,6)<br> no solution

kirill115 [55]

Answer:

x = 4 , y = 6

Step-by-step explanation:

Solve the following system:

{x + 3 y = 22 | (equation 1)

2 x - y = 2 | (equation 2)

Swap equation 1 with equation 2:

{2 x - y = 2 | (equation 1)

x + 3 y = 22 | (equation 2)

Subtract 1/2 × (equation 1) from equation 2:

{2 x - y = 2 | (equation 1)

0 x+(7 y)/2 = 21 | (equation 2)

Multiply equation 2 by 2/7:

{2 x - y = 2 | (equation 1)

0 x+y = 6 | (equation 2)

Add equation 2 to equation 1:

{2 x+0 y = 8 | (equation 1)

0 x+y = 6 | (equation 2)

Divide equation 1 by 2:

{x+0 y = 4 | (equation 1)

0 x+y = 6 | (equation 2)

Collect results:

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7 0

3 years ago

Consider the differential equation y'' − y' − 20y = 0. Verify that the functions e−4x and e5x form a fundamental set of solution

KIM [24]

Answer:

Therefore $e^{-4x}$ and $e^{5x}$ are fundamental solution of the given differential equation.

Therefore $e^{-4x}$ and $e^{5x}$ are linearly independent, since $W(e^{-4x},e^{5x})=9e^x\neq 0$

The general solution of the differential equation is

$y=c_1e^{-4x}+c_2e^{5x}$

Step-by-step explanation:

Given differential equation is

y''-y'-20y =0

Here P(x)= -1, Q(x)= -20 and R(x)=0

Let trial solution be $y=e^{mx}$

Then, $y'=me^{mx}$ and $y''=m^2e^{mx}$

$\therefore m^2e^{mx}-m e^{mx}-20e^{mx}=0$

$\Rightarrow m^2-m-20=0$

$\Rightarrow m^2-5m+4m-20=0$

$\Rightarrow m(m-5)+4(m-5)=0$

$\Rightarrow (m-5)(m+4)=0$

$\Rightarrow m=-4,5$

Therefore the complementary function is = $c_1e^{-4x}+c_2e^{5x}$

Therefore $e^{-4x}$ and $e^{5x}$ are fundamental solution of the given differential equation.

If $y_1$ and $y_2$ are the fundamental solution of differential equation, then

$W(y_1,y_2)=\left|\begin{array}{cc}y_1&y_2\\y'_1&y'_2\end{array}\right|\neq 0$

Then $y_1$ and $y_2$ are linearly independent.

$W(e^{-4x},e^{5x})=\left|\begin{array}{cc}e^{-4x}&e^{5x}\\-4e^{-4x}&5e^{5x}\end{array}\right|$

$=e^{-4x}.5e^{5x}-e^{5x}.(-4e^{-4x})$

$=5e^x+4e^x$

$=9e^x\neq 0$

Therefore $e^{-4x}$ and $e^{5x}$ are linearly independent, since $W(e^{-4x},e^{5x})=9e^x\neq 0$

Let the the particular solution of the differential equation is

$y_p=v_1e^{-4x}+v_2e^{5x}$

$\therefore v_1=\int \frac{-y_2R(x)}{W(y_1,y_2)} dx$

and

$\therefore v_2=\int \frac{y_1R(x)}{W(y_1,y_2)} dx$

Here $y_1= e^{-4x}$ , $y_2=e^{5x}$ , $W(e^{-4x},e^{5x})=9e^x$ ,and $R(x)=0$

$\therefore v_1=\int \frac{-e^{5x}.0}{9e^x}dx$

=0

and

$\therefore v_2=\int \frac{e^{5x}.0}{9e^x}dx$

=0

The the P.I = 0

The general solution of the differential equation is

$y=c_1e^{-4x}+c_2e^{5x}$

7 0

3 years ago

Identify the terms and like terms in the expression.

Cloud [144]

Terms of an expression are any Numbers separated by a plus or minus sign so your terms would be t, 8, & 3t

8 0

3 years ago