Find the tenth term in the sequence

Mathematics

1 answer:

attashe74 [19]2 years ago

4 0

Answer: The tenth term in the sequence is 512.

Step-by-step explanation:

Since we have given that

$a_n=2^{n-1}$

We need to find the tenth term:

It means n = 10

So, it becomes

$a_{10}=2^{10-1}\\\\a_{10}=2^9\\\\a_{10}=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\\\\a_{10}=512$

Hence, the tenth term in the sequence is 512.

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malfutka [58]

Pay attention in math class! in higher grade levels, this is gonna show up EVERYDAY! no joke
In these types of problems, you need to use PEMDAS also.

((4)+3)+(13(-2)-5)
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8 0

3 years ago

Help and show work that will make my day ❤️

kolezko [41]

H equals 30 because 3 times 6 is 18 so 5 times 6 is 30

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3 years ago

PLEASE HELP!!!!!!!dgbdgdbhdndcn

bogdanovich [222]

Problem 1)

AC is only perpendicular to EF if angle ADE is 90 degrees

(angle ADE) + (angle DAE) + (angle AED) = 180
(angle ADE) + (44) + (48) = 180
(angle ADE) + 92 = 180
(angle ADE) + 92 - 92 = 180 - 92
angle ADE = 88

Since angle ADE is actually 88 degrees, we do NOT have a right angle so we do NOT have a right triangle

Triangle AED is acute (all 3 angles are less than 90 degrees)

So because angle ADE is NOT 90 degrees, this means AC is NOT perpendicular to EF

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Problem 2)

a) The center is (2,-3)

The center is (h,k) and we can see that h = 2 and k = -3. It might help to write (x-2)^2+(y+3)^2 = 9 into (x-2)^2+(y-(-3))^2 = 3^3 then compare it to (x-h)^2 + (y-k)^2 = r^2

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b) The radius is 3 and the diameter is 6

From part a), we have (x-2)^2+(y-(-3))^2 = 3^3 matching (x-h)^2 + (y-k)^2 = r^2

where
h = 2
k = -3
r = 3

so, radius = r = 3
diameter = d = 2*r = 2*3 = 6

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c) The graph is shown in the image attachment. It is a circle with center point C = (2,-3) and radius r = 3.

Some points on the circle are

A = (2, 0)
B = (5, -3)
D = (2, -6)
E = (-1, -3)

Note how the distance from the center C to some point on the circle, say point B, is 3 units. In other words segment BC = 3.