Is the sequence an a solution of the recurrence relation an = 8an−1 − 16an−2?

Mathematics

1 answer:

rjkz [21]3 years ago

6 0

Hope this can help you

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15 - (-3) - 4 -16 22 -8 14

uranmaximum [27]

In this equation, you have to treat the number in the bracket first on the basis of BODMAS
15 - [-3]- 4
Note that when two minuses come together the product is a plus sign.
15 +3 - 4
You have to add before you subract
18 - 4 =14
Therefore, 15- [-3] - 4 = 14.

8 0

3 years ago

What is y intercept of the continuous in the table?

3241004551 [841]

I’m pretty sure it’s Number 3. Hope I’m Right.

4 0

3 years ago

Determine if the sequence is arithmetic. If it is, find the common difference. Is the sequence a function ?

BaLLatris [955]

$6)\\r=a_{n+1}-a_n\to r=a_2-a_1=a_3-a_2=a_4-a_3=...\\\\a_1=-7;\ a_2=-10;\ a_3=-11;\ a_4=-13\\\\a_2-a_1=-10-(-7)=-10+7=-3\\a_3-a_2=-11-(-10)=-11+10=-1\\\\a_3-a_2\neq a_2-a_1\\\\\text{It's not an arithmetic sequence}$

$\r=\dfrac{a_{n+1}}{a_n}\\\\r=\dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}=\dfrac{a_4}{a_3}=...\\\\8.)\\a_1=4;\ a_2=20;\ a_3=100;\ a_4=500\\\\r=\dfrac{20}{4}=\dfrac{100}{20}=\dfrac{500}{100}=5\\\\10.)\\a_1-30;\ a_2=-45;\ a_3=-50;\ a_4=-65\\\\\dfrac{a_2}{a_1}=\dfrac{-45}{-30}=\dfrac{3}{2}\\\\\dfrac{a_3}{a_2}=\dfrac{-50}{-45}=\dfrac{10}{9}\\\\\dfrac{a_2}{a_1}\neq\dfrac{a_3}{a_2}\\\\\text{It's not a geometric sequence}$
$12.)\\a_1=-7;\ a_2=-14;\ a_3=-21;\ a_4=-28\\\\\dfrac{a_2}{a_1}=\dfrac{-14}{-7}=2\\\\\dfrac{a_3}{a_2}=\dfrac{-21}{-14}=\dfrac{3}{2}\\\\\dfrac{a_2}{a_1}\neq\dfrac{a_3}{a_2}\\\\\text{It's not a geometric sequence}\\\\\text{It's a function}$

8 0

3 years ago

Pls show me proof, thank you

Karo-lina-s [1.5K]

The y has to be negative but the x positive. So that leaves only (3,-4)

4 0

3 years ago

II. Using Radians to Measure Arcs and Angles

padilas [110]

Pi radians is 180 degrees.
A
1) 180/3=60
2) 270
3) 45
4) 3240
B
1) 100/180=5/9
3) 30/180=3/18
5) 1/180
C
1) (pi) cm
2) 2pi

4 0

3 years ago