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3 years ago

What is this question?

Mathematics

1 answer:

Ira Lisetskai [31]3 years ago

7 0

Albert because 1/2 or .5 is grater than .4

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Answers must be in standard form:<br> (K+4)(3k+2)

kotykmax [81]

Answer:

3kK+2K+12k+8

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%5Csf%2015%3D2x-9" id="TexFormula1" title="\sf 15=2x-9" alt="\sf 15=2x-9" align="absmiddle" cl

Genrish500 [490]

Answer:

x=12

Step by step explanation:

Step by step explanation:Add 9 to both sides :

15 = 2x - 9

15 + 9 = 2x - 9 + 9

Simplify:

24 = 2x

Divide both sides to get the result:

24 = 2x

24/2 = (2x)/2

Result:

x=12

$\boxed{\sf Answered \: By\: Subhash}$

$\boxed{\sf Good\: luck \: for \: your \:assignment}$

7 0

2 years ago

Given the graph of g(x), describes the transformation of the parent function f(x)=2^x

salantis [7]

Answer:

The transformation being described is from g(x)=x2 g ( x ) = x 2 to f(x)=x2 f ( x ) = x 2 . The horizontal shift depends on the value of h . The horizontal shift is described as: f(x)=f(x+h) f ( x ) = f ( x + h ) - The graph is shifted to the left h units

6 0

2 years ago

3y''-6y'+6y=e*x sexcx

Simora [160]

From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

$3r^2-6r+6=0\iff r^2-2r+2=0$

which has roots at $r=1\pm i$ . This admits the two fundamental solutions

$y_1=e^x\cos x$
$y_2=e^x\sin x$

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

$y_p=u_1y_1+u_2y_2$

where

$u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$

and $W(y_1,y_2)$ is the Wronskian of the fundamental solutions. We have

$W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}$

and so

$u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx$
$u_1=\dfrac13\ln|\cos x|$

$u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx$
$u_2=\dfrac13x$

Therefore the particular solution is

$y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

so that the general solution to the ODE is

$y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

7 0

3 years ago

On a coordinate plane, a line goes through (negative 3, negative 3) and (negative 1, 5).

slega [8]

Answer:

y=20

x=4

Step-by-step explanation:

Since you can plug the first 2 coordinates into the slope equation ( $\frac{y_{2}-y_{1}}{x_{2} -x_{1} }$ ) you will get 4 for the slope so for the equation for the parallel line you would get y=4x+4 and since the x=4 you could plug that in and you would get y=20 and x=4

5 0

3 years ago

Read 2 more answers