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3 years ago

In the box, type in the number that will correctly complete the sentence.

Mathematics

2 answers:

kirill115 [55]3 years ago

8 0

She had 3 pencils left for herself. Hope this helps my dude ^^

Airida [17]3 years ago

7 0

If she have 12 pencils and she gave 3/4 away then she will have 3 pencils left.

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which of the following expressions does not contain any like terms? a. 2x^2+3x-4x+2 b. xy+yz-17 c. x^3+x^2-2x-x^2 d. 14+8+y

solniwko [45]

Answer : B

The expression xy+yz-17 do not contain any like terms

I hope that's help !

4 0

3 years ago

A shuttle launch depends on three key devices that may fail independently of each other with probabilities 0.01, 0.02, and 0.02,

worty [1.4K]

Answer:

$0.950796$

Step-by-step explanation:

Given that a shuttle launch depends on three key devices that may fail independently of each other with probabilities 0.01, 0.02, and 0.02, respectively.

Required probability = the probability for the shuttle to be launched on time

= Probability that all three do not fail

Since each key device is independent of the other

we have

prob that all three do not fail = $(1-0.01)(1-0.02)(1-0.02)\\=0.99*0.98*0.98\\=0.950796$

6 0

3 years ago

Construct perpendiculars image below

Marysya12 [62]

Answer: draw a straight line trough point B, same thing with the second one,for the third you must draw a straight line from the angle across to the segment. (make sure all of the intersections are 90 degrees

5 0

3 years ago

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 <em>the Green's Theorem </em>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

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

4 0

3 years ago

Whats the answer to this question

Liono4ka [1.6K]

Answer:

I believe it would be B.

Step-by-step explanation:

Range: area of variation with real numbers

Also, seeing the quadrant it is in means there are negative numbers. This takes out the possibility of option A.

4 0

3 years ago