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3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Mathematics

1 answer:

777dan777 [17]3 years ago

5 0

Answer: (C) 13,120

<u>Step-by-step explanation:</u>

Given the sequence {4, 12, 36, 108, ... , 8748} we know that the first term (a) is 4 and the ratio (r) is $\dfrac{12}{4}=3$

Input the values above into the Sum formula:

$S_n=\dfrac{a(1-r^n)}{1-r}\\\\S_8=\dfrac{4(1-3^8)}{1-3}\\\\.\quad=\dfrac{4(1-6561)}{-2}\\\\.\quad=\dfrac{4(-6560)}{-2}\\\\.\quad=-2(-6560)\\\\.\quad=13,120$

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Witch of the following could not represent the probability of an event

Alex17521 [72]

I think it's D. 7 times out of 34 times

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3 years ago

Which ordered pairs are solutions to the inequality 2x-y>1?

RideAnS [48]

Answer:

Any points in the shaded region including (2,-2) and (-3,-8)

Step-by-step explanation:

Convert the line into slope intercept form and graph it.

2x-y > 1 becomes -y>1-2x. Divide both sides by -1 and you get y<2x-1. Graph it with the shaded area on the right and a dashed line.

Any point which falls within the shaded red of the graph is a solution. No points on the line since it is not equal to (its dashed) are solutions. Check the location of your points to verify that they fall within this area.

(-3, -8) ---Yes

(-1, -3) ---No

(0, 5) --- No

(1, 6) --- No

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

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

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

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