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3 years ago

An arithmetic sequence begins as follows: a1=13 a2=19 Which of the following gives the definition of its nth term?

Mathematics

1 answer:

Otrada [13]3 years ago

5 0

Answer:

the nth term of the sequence is $a_n=6n+7$

Step-by-step explanation:

Given : An arithmetic sequence begins as follows: $a_1=13, a_2=19$

To find : Which of the following gives the definition of its nth term?

Solution :

The nth term of the A.P is $a_n=a+(n-1)d$

The first term is $a=a_1=13$

The common difference is $d=a_2-a_1$

$d=19-13=6$

Substitute in the formula,

$a_n=13+(n-1)6$

$a_n=13+6n-6$

$a_n=6n+7$

Therefore, the nth term of the sequence is $a_n=6n+7$

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What is the simplified form of (4x^2)^3

barxatty [35]

Answer:

i guess the answer is 64x^5

Step-by-step explanation:

(4x^2)^3

=64x^5

7 0

2 years ago

Which of the following geometric series converges?

Artist 52 [7]

All three series converge, so the answer is D.

The common ratios for each sequence are (I) -1/9, (II) -1/10, and (III) -1/3.

Consider a geometric sequence with the first term a and common ratio |r| < 1. Then the n-th partial sum (the sum of the first n terms) of the sequence is

$S_n=a+ar+ar^2+\cdots+ar^{n-2}+ar^{n-1}$

Multiply both sides by r :

$rS_n=ar+ar^2+ar^3+\cdots+ar^{n-1}+ar^n$

Subtract the latter sum from the first, which eliminates all but the first and last terms:

$S_n-rS_n=a-ar^n$

Solve for $S_n$ :

$(1-r)S_n=a(1-r^n)\implies S_n=\dfrac a{1-r}-\dfrac{ar^n}{1-r}$

Then as gets arbitrarily large, the term $r^n$ will converge to 0, leaving us with

$S=\displaystyle\lim_{n\to\infty}S_n=\frac a{1-r}$

So the given series converge to

(I) -243/(1 + 1/9) = -2187/10

(II) -1.1/(1 + 1/10) = -1

(III) 27/(1 + 1/3) = 18

8 0

3 years ago

The shape of the clock is a regular dodecagon with a radius of 14cm. Centered on the clock's face is a green circle of radius 9c

faust18 [17]

The area of a dodecagon is 1/2aP
where:
a-apothem
P-perimeter

In the figure you will draw a point to any side and make a line perpendicular from a center to that side. Then if u think u can see that with only one draw of a hypotenuse can form a triangle. Then after drawing the hypotenuse which has 2 triangles, make the top angle of 1 triangle there are 30 degrees. On the down part is perpendicular and the left or the right is the 60 degrees. After that using special right triangles like the 30, 60, 90,. Then at the base of the one it bisected dived them by 2 for the 2 triangle's base. After that solve using it since the radius is 14. The apothem is 14(sqrt of 3). So after solving them the perimeter of each dodecagon is 12 sided so 14x12 is 168

A=1/2aP
A=1/2(14)(sqrt of 3)(168)
A=1176(sqrt of 3)

Thats the area of the dodecagon now the circle

A=(pi)r^2
A=(pi)(9)^2
A=81pi

Therefore the color purple is greater than the color green

8 0

3 years ago

Please help me with these problems :( -random person who struggles in pre algebra

yKpoI14uk [10]

Hope these help:
#9
a-4 divided by 5 = 12
5*12 is 60
60+4 is 64
60 (final answer for a)
#10
n+3 divided by 8 = -4
8*(-4) = -32
-32 + 3 = -35
-32 (final answer for n)
#11
6+z divided by 10 = -2
-2 * 10 = -20
-20=6=-26
-20 (final answer for z)
btw, can't see diagram

8 0

3 years ago

What is the sum (Two-fifths x + StartFraction 5 over 8 EndFraction) + (one-fifth x minus one-fourth)?

liubo4ka [24]

The sum of ( $\frac{2}{5}$ x + $\frac{5}{8}$ ) + ( $\frac{1}{5}$ x - $\frac{1}{4}$ ) is $\frac{3}{5}$ x + $\frac{3}{8}$

Step-by-step explanation:

To add or subtract 2 fractions, they must have same denominators, if they not do that

Find the Lowest common multiple of the two denominators (LCM)
Replace each denominator by it
Divide The LCM by each denominator and multiply the numerator of each fraction by its quotient

∵ ( $\frac{2}{5}$ x + $\frac{5}{8}$ ) + ( $\frac{1}{5}$ x - $\frac{1}{4}$ )

- Let us start with adding the like terms

∴ ( $\frac{2}{5}$ x + $\frac{5}{8}$ ) + ( $\frac{1}{5}$ x - $\frac{1}{4}$ ) = (

∵ ( $\frac{2}{5}$ x + $\frac{1}{5}$ x ) have same denominators, then we can add them

∴ ( $\frac{2}{5}$ x + $\frac{1}{5}$ x ) = $\frac{3}{5}$ x

∵ ( $\frac{5}{8}$ - $\frac{1}{4}$ ) do not have the same denominators, then we must find

LCM of 8 and 4

- The LCM of 8 and 4 is 8 because 8 is the first common multiple

of 8 and 4

∵ LCM of 8 and 4 is 8

- Divide 8 by the denominator 4

∵ 8 ÷ 4 = 2

- Multiply the numerator of the fraction by 2

∴ $\frac{1}{4}=\frac{2}{8}$

∴ ( $\frac{5}{8}$ - $\frac{1}{4}$ ) = ( $\frac{5}{8}$ - $\frac{2}{8}$ ) = $\frac{3}{8}$

∴ ( $\frac{2}{5}$ x + $\frac{5}{8}$ ) + ( $\frac{1}{5}$ x - $\frac{1}{4}$ ) = $\frac{3}{5}$ x + $\frac{3}{8}$

The sum of ( $\frac{2}{5}$ x + $\frac{5}{8}$ ) + ( $\frac{1}{5}$ x - $\frac{1}{4}$ ) is $\frac{3}{5}$ x + $\frac{3}{8}$

Learn more:

You can learn more about the fractions in brainly.com/question/2456302

#LearnwithBrainly

7 0

3 years ago

Read 2 more answers