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2 years ago

How can I factorise expressions

Mathematics

1 answer:

expeople1 [14]2 years ago

7 0

Answer:

The first step of factorising an expression is to 'take out' any common factors which the terms have. So if you were asked to factorise x² + x, since x goes into both terms, you would write x(x + 1) . This video shows you how to solve a quadratic equation by factoring.

Step-by-step explanation:

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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

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

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2 years ago

For one month Siera calculated her home town’s average high temperature in degrees Fahrenheit. She wants to convert that tempera

pishuonlain [190]

C(F) is the degrees Celsius given the degrees Fahrenheit.<span />

4 0

3 years ago

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Which of the following represents the factorization of the trinomal below

Allisa [31]

I believe B. is your answer. If you use the ac method, you will multiply 8 and 3 which is 24. So now, what factors of 24 will give you 10?6 and 4

Now substitute

$8 {x}^{2} + 4x + 6x + 3$
Now do factorization by grouping.

$4x(2x + 1) + 3(2x + 1)$
Therefore, your answer is (4x+3)2x+1)

6 0

3 years ago

15y^-11/3y^-11, show steps I need help please don't understand how to do it.

tester [92]

One may note, you never quite asked anything per se, ahemm, if you meant simplification.

$\bf \left.\qquad \qquad \right.\textit{negative exponents}\\\\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}
\qquad \qquad
\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\qquad \qquad 
a^{{{ n}}}\implies \cfrac{1}{a^{-{{ n}}}}\\\\
-------------------------------\\\\
\cfrac{15y^{-11}}{3y^{-11}}\implies \cfrac{15}{3}\cdot \cfrac{y^{-11}}{y^{-11}}\implies 5\cdot y^{-11}\cdot y^{+11}\implies 5\cdot y^{-11+11}
\\\\\\
5y^0 \implies \boxed{5}$

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3 years ago

Find angle a. <br> You will get 20 points

VARVARA [1.3K]

Answer:

127

Step-by-step explanation:

15+38+a=180

a= 127

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2 years ago

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