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Mathematics

2 answers:

skelet666 [1.2K]3 years ago

8 0

Your correct dude keep trying ;)

yawa3891 [41]3 years ago

6 0

Answer:

1) Not a function 2) Function 3) Not a function 4) Not a function 5) Function

Step-by-step explanation:

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Space Shuttle has three computers. Computer A is a primary computer and Computer B and Computer C are auxiliary computers. There

sweet-ann [11.9K]

Answer:

Required probability equals 0.18%

Step-by-step explanation:

The probability that the primary and one auxiliary computer fails equals

1) Probability that A and B fails

2)Probability that A and C fails

Thus required probability equals

$P(E)=P(1)(2)+P(1)P(3)\\\\P(E)=0.03\times 0.03+0.03\times 0.03P(E)=0.18%$

3 0

3 years ago

Divide mentally 270 divide 3=

yanalaym [24]

27/3 = 9 so 270/3 = 90

4 0

3 years ago

Read 2 more answers

Let F⃗ =2(x+y)i⃗ +8sin(y)j⃗ .

Alik [6]

Answer:

-42

Step-by-step explanation:

The objective is to find the line integral of $F$ around the perimeter of the rectangle with corners (4,0), (4,3), (−3,3), (−3,0), traversed in that order.

We will use the Green's Theorem to evaluate this integral. The rectangle is presented below.

We have that

$F(x,y) = 2(x+y)i + 8j \sin y = \langle 2(x+y), 8\sin y \rangle$

Therefore,

$P(x,y) = 2(x+y) \quad \wedge \quad Q(x,y) = 8\sin y$

Let's calculate the needed partial derivatives.

$P_y = \frac{\partial P}{\partial y} (x,y) = (2(x+y))'_y = 2\\Q_x =\frac{\partial Q}{\partial x} (x,y) = (8\sin y)'_x = 0$

Thus,

$Q_x -P_y = 0 -2 = - 2$

Now, by the Green's theorem, we have

$\oint_C F \,dr = \iint_D (Q_x-P_y)\,dA = \int \limits_{-3}^{4} \int \limits_{0}^{3} (-2)\,dy\, dx \\ \\\phantom{\oint_C F \,dr = \iint_D (Q_x-P_y)\,dA}= \int \limits_{-3}^{4} (-2y) \Big|_{0}^{3} \; dx\\ \phantom{\oint_C F \,dr = \iint_D (Q_x-P_y)\,dA}= \int \limits_{-3}^{4} (-6)\; dx = -6x \Big|_{-3}^{4} = -42$