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REY [17]
3 years ago
15

Find the 7th term of the geometric sequence with the given terms.

Mathematics
1 answer:
ohaa [14]3 years ago
3 0

Answer:

1. 1458

2. 500 or -500

Step-by-step explanation:

1. a4=54, a5=162

r = 162/54 = 3

a7 = a5 × r²

= 162 × 3² = 1458

2.a4=-4, a6=-100

r² = -100/-4 = 25

r = +/- 5

a7 = a6 × r

= -100 × +/- 5

= +/- 500

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Geometry is basically shapes. It includes area, volume, and surface area.
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A bacteria cell splits into two cells every hour. Write and evaluate an exponential expression to find how many cells there will
ryzh [129]
Easy

equation is
f(x)=2ˣ

where x=number of hours elapsed

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to solve for 1024

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using calculator
10=x

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4 0
3 years ago
These base-10 blocks model 68.
Rina8888 [55]

Answer:

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−      

7 0
2 years ago
Determine as a linear relation in x, y, z the plane given by the vector function F(u, v) = a + u b + v c when a = 2 i − 2 j + k,
Ostrovityanka [42]

Answer:

2x - y - 3z = 0

Step-by-step explanation:

Since the set

{i, j}  = {(1,0), (0,1)}

is a base in \mathbb{R}^2

and F is linear, then

<em>{F(1,0), F(0,1)}  </em>

would be a base of the plane generated by F.

F(1,0) = a+b = (2i-2j+k)+(i+2j+k) = 3i+2k

F(0,1) = a+c = (2i-2j+k)+(2i+j+2k) = 4i-j+3k

Now, we just have to find the equation of the plane that contains the vectors 3i+2k and 4i-j+3k

We need a normal vector which is the cross product of 3i+2k and 4i-j+3k

(3i+2k)X(4i-j+3k) = 2i-j-3k

The equation of the plane whose normal vector is 2i-j-3k and contains the point (3,0,2) (the end of the vector F(1,0)) is given by

2(x-3) -1(y-0) -3(z-2) = 0

or what is the same

2x - y - 3z = 0

3 0
3 years ago
Round your answer to this problem to the nearest degree.
Vika [28.1K]
Use the sine law: sinB/b = sinA/a

SinB = b*sinA/a

sinB = 3*sin120° /8

∠ B = 18.94° ≈ 19°
5 0
3 years ago
Read 2 more answers
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