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DerKrebs [107]
3 years ago
10

What is the common difference or common ratio for the sequence: 27, 9, 3, 1, 1/3, ...

Mathematics
1 answer:
lawyer [7]3 years ago
4 0

Answer:

It's dividing by 3 everytime, so 1/3

Step-by-step explanation:

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Neeeddddddd help :/.......
Angelina_Jolie [31]

Answer:

a

Step-by-step explanation:

7 0
2 years ago
The number of seconds varies directly as the number of minutes. When 300 seconds have passes, 5 minutes have passed. If 480 seco
Stells [14]
Hello!

We know that 1 minute = 60 seconds.

5 minutes = 300 seconds because 60 × 5 = 300

360 seconds would be 6 muinutes. 420 seconds would be 7 minutes. That means 480 seconds = 8 minutes, so 8 minutes have passed.
5 0
2 years ago
Prove that the two circles shown below are similar.
s344n2d4d5 [400]

The circles B and D are similar since the radius of the former is as twice as the radius of the latter.

<h3>Are the two circles shown in the figure similar?</h3>

By geometry we know that circles are defined by only one characteristic: radius. If two circles are similar, then the radii must be <em>different</em>. After a quick look at the figure, we conclude that the radii are r_{B} = 4 and r_{D} = 2. Hence, the circles B and D are similar since the radius of the former is as twice as the radius of the latter.

To learn more on similarity: brainly.com/question/12670209

#SPJ1

5 0
1 year ago
What is the sum of the first 7 terms of the series −4+8−16+32−... ?
pochemuha

we have that

−4+8−16+32−.....

a1=-2*(-2)-----> -4

a2=-4*(-2)-----> +8

a3=+8*(-2)-----> -16

a4=-16*(-2)----> +32

a5=+32*(-2)----> -64

a6=-64*(-2)-----> +128

a7=+128*(-2)-----> -256

The sum of the first 7 terms of the series is

<span>[a1+a2+a3+a4+a5+a6+a7]-----> [-4+8-16+32-64+128-256]-------> -172</span>

<span>
the answer is -172</span>
4 0
2 years ago
How do find the general solution of sin7A=sinA+sin3A
const2013 [10]

Step-by-step answer:

Solve  

sin(7a) = sin(3a) + sin(a) ..................(1)

Let  

F(a)=sin(7a)-sin(3a)-sin(a)..................(2)

the equivalent problem to (1) is  

F(a)=0 ......................................(3)

F(a)

=sin(7a)-sin(3a)-sin(a)....apply trig sum identities

=sin(6a+a) - sin(3a) - sin(a)

=sin(6a)cos(a)+cos(6a)sin(a) -sin(3a) - sin(a)

apply double angle formulas

=(2sin(3a)cos(3a))cos(a) +

(cos^2(3a)-sin^2(3a))sin(a) -sin(3a) - sin(a)

simplify using sin^2(p)+cos^2(p) = 1

= (2sin(3a)cos(3a))cos(a) +

(1-2sin^2(3a))sin(a) - sin(3a) - sin(a)

simplify algebraically, note 1*sin(a) cancels sin(a)

= (2sin(3a)cos(3a))cos(a) -2sin^2(3a)sin(a) - sin(3a)

factor out sin(3a)

= sin(3a)(2cos(3a)cos(a)-2sin(3a)sin(a) - 1)

now use trig sum formula to reduce to cos(4a)

= sin(3a)(2cos(4a)-1)

So

F(a) = 0   if sin(3a) =0 ...................(4)

or

F(a) = 0   if cos(4a) = 1/2 ................(5)

using the zero product theorem

From (4)

sin(3a) = 0  

3a = sin-1(0) = n*pi

a = n*pi/3  ................................(6)

From (5)

cos(4a) = 1/2

4a = cos^-1(1/2) = 2n*pi +/- pi/3

a = (2n+1/3)pi/4 or (2n-1/3)pi/4............(7)

Combine (6) and (7) to give the general solution

a = n*pi/3 or (2n+1/3)pi/4 or (2n-1/3)pi/4 .....(8)

For checking, use your calculator to substitute every solution given in (8) into the function F(a) in equation (1) to confirm that the result is almost equality.  The small difference will be due to rounding errors of the calculators.  On mine, they work out to be of the order of +/- 10^-15.

6 0
3 years ago
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