2 years ago

Find the 15th term of the geometric sequence 2 , -8 , 32

Mathematics

1 answer:

marishachu [46]2 years ago

5 0

Answer:

first term (a)= 2

second term (b) = -8

common ratio (r) = -8/2 = -4

15th term = ar^ (n-1)

= 2* (-4)^ 14 =536870912

Step-by-step explanation:

You might be interested in

Last month there was 4 sunny days for every raining day

anzhelika [568]

Answer:

Step-by-step explanation:

8 0

2 years ago

Enter the numbers that goes in the boxes.

sergey [27]

Answer:

A'(-1,4)

B'(-4,3)

C'(-2,1)

6 0

3 years ago

(√2+√3)² + (√5+√2)²

harkovskaia [24]

Answer:

12+2√6+2√10

Step-by-step explanation:

(√2+√3)² + (√5+√2)²=2+3+2√6+5+2+2√10=12+2√6+2√10

6 0

3 years ago

Read 2 more answers

pls help me with this if you know it also to explain it thanks Solve the equation -21 + 70n = 77 for n.

gogolik [260]

Answer:

n = 1 2/5

Step-by-step explanation:

-21 + 70n = 77

Add 21 to each side

-21+21 + 70n = 77 +21

70n = 98

Divide each side by 70

70n/70 = 98/70

n =98/70

n = 70/70 +28/70

n = 1 + 2/5

n = 1 2/5

6 0

3 years ago

Find in degrees the numeric value of the acute angle of rotation that eliminates the product term from the equation 7x2+24xy−34x

lutik1710 [3]

Rotating the graph of $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ with $B\ne0$ counterclockwise by $\theta$ eliminates the $xy$ term, where $\cot2\theta=\frac{A-C}{B}.$ Plugging in, we have $\cot2\theta=\frac{7}{24},$ since $C=0.$ Solving, we have $\theta=\frac{1}{2}\cot^{-1}(\frac{7}{24})\approx\boxed{36.9^\circ}.$

4 0

3 years ago