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3 years ago

Sorry another question!

Mathematics

1 answer:

Julli [10]3 years ago

8 0

Answer:

you start counting from left to right

when you want to calculate the difference you have to choose two numbers that are next to each other and subtract second one from first one and because the different between every 2 numbers in order are the same numbers we call it a common difference.

Step-by-step explanation:

for example for 4 :

1. we can choose two numbers next to each other like 1 and 1.1

2. subtract second from first like 1.1 - 1 = 0.1

3. 1.1 + 0.1 = 1.2

1.2 + 0.1 = 1.3

and ...

so the common difference is 0.1

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Suppose a white dwarf star has a diameter of approximately 1.8083 to the power of 4 km. Use the formula 4n to the power of 2 to

aivan3 [116]

ANSWER:

The surface area of the star is 3.2700 x $10^{8}$ square kilometres.

EXPLANATIONS:

Diameter of the star = 1.8083 x $10^{4}$ Km.

Surface area of the star = 4 $n^{2}$

Where n is the radius of the star.

So that;

n = $\frac{ 1.8083 * 10^{4} }{2}$

= 0.90415 x $10^{4}$

n = 0.90415 x $10^{4}$ Km

Thus,

Surface area = 4 x $(0.90415*10^{4} )^{2}$

= 326994889

Surface area = 3.2700 x $10^{8}$ $km^{2}$

Therefore, the surface area of the star is 3.2700 x $10^{8}$ square kilometres.

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

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

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

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