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zavuch27 [327]
3 years ago
15

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&gt;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

(2, -2) ---Yes

8 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
Use distributive property to clear parentheses-(3x-4y+5z)=
Vadim26 [7]

Answer:

-3x + 4y -5z

Step-by-step explanation:

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6 0
3 years ago
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bearhunter [10]
Because you add both negative x’s which is -2 and add the -6 in the right side so then it’s -2x=4
Divide and you get x=-2
6 0
3 years ago
Read 2 more answers
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