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3 years ago

Write down the next term in this sequence: 3 8 15 24 ...

Mathematics

2 answers:

Sav [38]3 years ago

8 0

Answer:

there 2 ways to answer this, you could find the nth term first or you could add the amount between them.

1st way-

3 8 15 24

8-3 = 5 so this is like the 5 times table

5 10 15 20

nth term- 5n+4

so 5x5+4=29

2nd way-

3-8= 5

24+5= 29

if this was for a test, i would do this first answer as you would get more marks for the working out

hope this helps

(sorry is it is wrong)

Zarrin [17]3 years ago

4 0

35 because the in the second line of the pic there is a difference of 2

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There are 95 white pieces of paper in a hat along with 5 blue pieces of paper. What is the theoretical probability that a blind-

sp2606 [1]

Answer:

Step-by-step explanation:

Probability is the likelihood of an event taking place.

-Given there are 95 white and 5 blue pieces, the total number of papers is 100.

-The theoretical first attempt probability is equivalent to the number of each paper divided by the total:

$P(blue)= \frac{no \ of \ blue}{total}\\\\=\frac{5}{100}\\\\=0.05$

Hence, the theoretical probability that a blind-folded person draws a blue piece of paper on the first attempt is 0.05 or 5%

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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 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

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