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Ulleksa [173]
3 years ago
11

What is the common ratio of the geometric sequence below ? -96 , 48 , -24 , 12 , -6

Mathematics
2 answers:
UkoKoshka [18]3 years ago
8 0
The common ratio is -1/2

-96 * -1/2 = 48
48 * -1/2 = -24
-24 * -1/2 = 12
12 * -1/2 = 6
SCORPION-xisa [38]3 years ago
5 0
The numbers are in the order of
Negative, positive, negative, positive, negative
Or
It starts with a negative number, ends with a negative.
This is a pattern either way.
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-7x - 2y = 19<br> 4x + y = -12<br> Solve the system
nalin [4]

-7x - 2y = 19

4x + y = -12

Set y equal to each other (opposite signs are fine and you could also set x equal instead of y)

-7x - 2y = 19

8x + 2y = -76

Add equations together

x = -52

Plug x value into an equation

4(-52) + y = -12

Solve for y

-208 + y = -12

y = 196

Hope this helps! ;)

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Answer:

Hi! Your answer is down below!

Step-by-step explanation:

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Find constants a and b such that the function y = a sin(x) + b cos(x) satisfies the differential equation y'' + y' − 5y = sin(x)
vichka [17]

Answers:

a = -6/37

b = -1/37

============================================================

Explanation:

Let's start things off by computing the derivatives we'll need

y = a\sin(x) + b\cos(x)\\\\y' = a\cos(x) - b\sin(x)\\\\y'' = -a\sin(x) - b\cos(x)\\\\

Apply substitution to get

y'' + y' - 5y = \sin(x)\\\\\left(-a\sin(x) - b\cos(x)\right) + \left(a\cos(x) - b\sin(x)\right) - 5\left(a\sin(x) + b\cos(x)\right) = \sin(x)\\\\-a\sin(x) - b\cos(x) + a\cos(x) - b\sin(x) - 5a\sin(x) - 5b\cos(x) = \sin(x)\\\\\left(-a\sin(x) - b\sin(x) - 5a\sin(x)\right)  + \left(- b\cos(x) + a\cos(x) - 5b\cos(x)\right) = \sin(x)\\\\\left(-a - b - 5a\right)\sin(x)  + \left(- b + a - 5b\right)\cos(x) = \sin(x)\\\\\left(-6a - b\right)\sin(x)  + \left(a - 6b\right)\cos(x) = \sin(x)\\\\

I've factored things in such a way that we have something in the form Msin(x) + Ncos(x), where M and N are coefficients based on the constants a,b.

The right hand side is simply sin(x). So we want that cos(x) term to go away. To do so, we need the coefficient (a-6b) in front of that cosine to be zero

a-6b = 0

a = 6b

At the same time, we want the (-6a-b)sin(x) term to have its coefficient be 1. That way we simplify the left hand side to sin(x)

-6a  -b = 1

-6(6b) - b = 1 .... plug in a = 6b

-36b - b = 1

-37b = 1

b = -1/37

Use this to find 'a'

a = 6b

a = 6(-1/37)

a = -6/37

8 0
3 years ago
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