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1 year ago

Evaluate the expression 2x-(4y-3)+5xz, when x=-3,y=2, and z=-1

Mathematics

1 answer:

AVprozaik [17]1 year ago

3 0

Answer: 4

Step-by-step explanation:

2x - (4y - 3) + 5xz

Let x = -3, y = 2, and z = ‐1.

2(-3) -( 4(2) -3) +5(-3)(-1)

-6 - (8-3) +15

-6 - 5 +15

-11 +15

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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

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

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

There are 12 students in a classroom, including the triplets joey, chloe, and zoe. If 3 of the 12 are randomly selcted to give s

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3 years ago

Read 2 more answers

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ladessa [460]

Answer:

$n(A \cup \: B) = n(A) + n(B) - n(A \cap \: B)\\ = 9 + 11 - 5 = 15 \\$

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2 years ago

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earnstyle [38]

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2 years ago

Find the product and simplify your answer. 7a5(-3a4+5a-5)

Alex Ar [27]

$\text{Hey there!}$

$\mathsf{Use\ distributive\ property\ to\ solve}}\\\\\text{The formula is : a(b + c)=a(b)+a(c)=b+ac}\\\\\mathsf{7a^5(-3a^4+5a-5)}\\\\\mathsf{7a^5(-3a^4)=-21a^9}\\\\\mathsf{7a^5(5a)=3a^6}\\\\\mathsf{7a^5(-5)=-35a^5}$

$\boxed{\boxed{\mathsf{Your\ answer: \boxed{\mathsf{-21a^9+35a^6-35a^5}}}}}}\checkmark$

$\text{Good luck on your assignment and enjoy your day!}$

~ $\dfrac{\frak{LoveYourselfFirst}}{:)}$

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2 years ago