All categories

2 years ago

Write the next three terms of the following sequence:

Mathematics

1 answer:

Keith_Richards [23]2 years ago

7 0

Answer:

21... 27.... 38...

Step-by-step explanation:

Add prime numbers to the number in alternate.

4 + 3 = 7

5 + 5 = 10

7 + 7 = 14

10 + 11 = 21

14 + 13 = 27

21 + 17 = 38

You might be interested in

Which choices are common factors of the numerator and denominator of 18/24

Andre45 [30]

6/8, 3/4 hopes this helps

5 0

2 years ago

We need help . we think we got it but were not sure

maria [59]

Answer:

your right

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Which do I click for the answer?

beks73 [17]

You Should click :

B) The row categories are "Wears Glasses" and "Do Not Wear Glasses"

C) Two teachers do not wear glasses

E) Total of 6 teachers were polled

<u>Explanation:</u>

In a class there are 129 students, among that 32 wear glasses, 97 don't wear glasses.
There are 6 teachers in those 4 wear glasses and 2 do not wear glasses.
Totally 135 present. In those 36 wear and 99 don't wear glasses.

7 0

3 years ago

Three times a number increased by eight is no more than a number decreased by four

Westkost [7]

Answer:

x is no more than -6

Step-by-step explanation:

3 0

3 years ago

Solve the differential. This was in the 2016 VCE Specialist Maths Paper 1 and i'm a bit stuck

Nimfa-mama [501]

$\sqrt{2 - x^{2}} \cdot \frac{dy}{dx} = \frac{1}{2 - y}$
$\frac{dy}{dx} = \frac{1}{(2 - y)\sqrt{2 - x^{2}}}$

Now, isolate the variables, so you can integrate.
$(2 - y)dy = \frac{dx}{\sqrt{2 - x^{2}}}$
$\int (2 - y)\,dy = \int\frac{dx}{\sqrt{2 - x^{2}}}$
$2y - \frac{y^{2}}{2} = sin^{-1}\frac{x}{\sqrt{2}} + \frac{1}{2}C$

$4y - y^{2} = 2sin^{-1}\frac{x}{\sqrt{2}} + C$
$y^{2} - 4y = -2sin^{-1}\frac{x}{\sqrt{2}} - C$
$(y - 2)^{2} - 4 = -2sin^{-1}\frac{x}{\sqrt{2}} - C$
$(y - 2)^{2} = 4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C$

$y - 2 = \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}$
$y = 2 \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}$

At x = 1, y = 0.
$0 = 2 \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$
$-2 = \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$

$4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C > 0$
$\therefore 2 = \sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$

$4 = 4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C$
$0 = -2sin^{-1}\frac{1}{\sqrt{2}} - C$
$C = -2sin^{-1}\frac{1}{\sqrt{2}} = -2\frac{\pi}{4}$
$C = -\frac{\pi}{2}$

$\therefore y = 2 - \sqrt{4 + \frac{\pi}{2} - 2sin^{-1}\frac{x}{\sqrt{2}}}$

6 0

3 years ago