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AURORKA [14]
3 years ago
9

A rule for the nth term of 22,9,-4,-17,-30

Mathematics
1 answer:
fenix001 [56]3 years ago
4 0

Answer: In Exercises 21 and 22, describe and correct the error in writing a rule for the nth term of the arithmetic sequence 22, 9, −4, −17, −30, . . ..

3. 1, −1, −3, −5, −7, . . . 4.

5. 5, 8, 13, 20, 29, . . . 6.

7. 36, 18, 9, —9, —9, . . . 8. 24

9. —1, —3, 1, —5, —3, . . . 10. 24 42

12, 6, 0, −6, −12, . . .

3, 5, 9, 15, 23, . . .

81, 27, 9, 3, 1, . . .

—1 , —1 , —5 , —7 , —3 , . . . 62662

21.

22. ✗

11. WRITING EQUATIONS Write a rule for the arithmetic sequence with the given description.

a. The first term is −3 and each term is 6 less than the previous term.

b. The first term is 7 and each term is 5 more than the previous term.

12. WRITING Compare the terms of an arithmetic sequence when d > 0 to when d < 0.

In Exercises 13–20, write a rule for the nth term of the sequence. Then find a20. (See Example 2.)

The first term is 22 and the common difference is −13.

an =−13+(n−1)(22) an =−35+22n

13. 12, 20, 28, 36, . . . 14. 15. 51, 48, 45, 42, . . . 16.

7, 12, 17, 22, . . .

27. a =−5,d=−—1

17 2 21 2

86, 79, 72, 65, . . . 11 511

29. USING EQUATIONS One term of an arithmetic sequence is a8 = −13. The common difference

is −8. What is a rule for the nth term of the sequence?

○A an = 51 + 8n ○B an = 35 + 8n ○C an=51−8n ○D an=35−8n

17. −1, −—3, —3, 1, . . . 18. 19. 2.3, 1.5, 0.7, −0.1, . . . 20.

−2, −—4, −—2, —4, . . . 11.7, 10.8, 9.9, 9, . . .

422 Chapter 8 Sequences and Series

✗

Use a1 = 22 and d = −13. a =a +nd

In Exercises 23–28, write a rule for the nth term of th

Step-by-step explanation:

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Natali [406]

The amount that the credit union will finance is $24169.60

We have given that the cost of the car is $22,346. 16

and the credit union required 10% down payment

and sale tax=7.6%

The license and title charges are $125. 13.

We have to determine that the  amount that the credit union will

finance Now  we have sale tax 7.6% for $22,346. 16

Therefore, we get

\frac{7.6}{100} \times 22,346. 16=1698.31

<h3>What is the amount that the credit union finance ?</h3>

credit union finance=cost of car+sales tax+the license and title  charges.

=22346.16+1698.31+12.13\\=24169.60

Therefore, the amount that the credit union will finance is $24169.60

To learn more about the credit union will finance visit:

brainly.com/question/15641576

4 0
2 years ago
I would love if someone would help me
mart [117]
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I’m not very familiar with the square roots method but I believe you try and isolate the term that is squared.

4 0
3 years ago
2+X/9 = 5x-6/27 SOLVE.
oksano4ka [1.4K]

Answer:

x = 6

Step-by-step explanation:

First, we need to cross multiply on both sides, which gives us:

9 * (5x - 6) = 27 * (2 + x)

45x - 54 = 54 + 27x

Now, we want to isolate x on either side.

We can substract 54 from both sides:

(45x - 54 = 54 + 27x) - 54

45x - 108 = 27x

We then subtract 45 from both sides:

(45x - 108 = 27x) - 45

-108 = -18x

Finally, we divide both sides by -18:

(-108 = -18x) / -18

6 = x

6 0
2 years ago
How to solve this trig
n200080 [17]

Hi there!

To find the Trigonometric Equation, we have to isolate sin, cos, tan, etc. We are also given the interval [0,2π).

<u>F</u><u>i</u><u>r</u><u>s</u><u>t</u><u> </u><u>Q</u><u>u</u><u>e</u><u>s</u><u>t</u><u>i</u><u>o</u><u>n</u>

What we have to do is to isolate cos first.

\displaystyle  \large{ cos \theta =  -  \frac{1}{2} }

Then find the reference angle. As we know cos(π/3) equals 1/2. Therefore π/3 is our reference angle.

Since we know that cos is negative in Q2 and Q3. We will be using π + (ref. angle) for Q3. and π - (ref. angle) for Q2.

<u>F</u><u>i</u><u>n</u><u>d</u><u> </u><u>Q</u><u>2</u>

\displaystyle \large{ \pi -  \frac{ \pi}{3}  =  \frac{3 \pi}{3}  -  \frac{  \pi}{3} } \\  \displaystyle \large \boxed{ \frac{2 \pi}{3} }

<u>F</u><u>i</u><u>n</u><u>d</u><u> </u><u>Q</u><u>3</u>

<u>\displaystyle \large{ \pi  +   \frac{ \pi}{3}  =  \frac{3 \pi}{3}   +   \frac{  \pi}{3} } \\  \displaystyle \large \boxed{ \frac{4 \pi}{3} }</u>

Both values are apart of the interval. Hence,

\displaystyle \large \boxed{ \theta =  \frac{2 \pi}{3} , \frac{4 \pi}{3} }

<u>S</u><u>e</u><u>c</u><u>o</u><u>n</u><u>d</u><u> </u><u>Q</u><u>u</u><u>e</u><u>s</u><u>t</u><u>i</u><u>o</u><u>n</u>

Isolate sin(4 theta).

\displaystyle \large{sin 4 \theta =  -  \frac{1}{ \sqrt{2} } }

Rationalize the denominator.

\displaystyle \large{sin4 \theta =  -  \frac{ \sqrt{2} }{2} }

The problem here is 4 beside theta. What we are going to do is to expand the interval.

\displaystyle \large{0 \leqslant  \theta < 2 \pi}

Multiply whole by 4.

\displaystyle \large{0 \times 4 \leqslant  \theta \times 4 < 2 \pi \times 4} \\  \displaystyle \large \boxed{0 \leqslant 4 \theta < 8 \pi}

Then find the reference angle.

We know that sin(π/4) = √2/2. Hence π/4 is our reference angle.

sin is negative in Q3 and Q4. We use π + (ref. angle) for Q3 and 2π - (ref. angle for Q4.)

<u>F</u><u>i</u><u>n</u><u>d</u><u> </u><u>Q</u><u>3</u>

<u>\displaystyle \large{ \pi +  \frac{ \pi}{4}  =  \frac{ 4 \pi}{4}  +  \frac{ \pi}{4} } \\  \displaystyle \large \boxed{  \frac{5 \pi}{4} }</u>

<u>F</u><u>i</u><u>n</u><u>d</u><u> </u><u>Q</u><u>4</u>

\displaystyle \large{2 \pi -  \frac{ \pi}{4}  =  \frac{8 \pi}{4}  -  \frac{ \pi}{4} } \\  \displaystyle \large \boxed{ \frac{7 \pi}{4} }

Both values are in [0,2π). However, we exceed our interval to < 8π.

We will be using these following:-

\displaystyle \large{ \theta + 2 \pi k =  \theta \:  \:  \:  \:  \:  \sf{(k  \:  \: is \:  \: integer)}}

Hence:-

<u>F</u><u>o</u><u>r</u><u> </u><u>Q</u><u>3</u>

\displaystyle \large{ \frac{5 \pi}{4}  + 2 \pi =  \frac{13 \pi}{4} } \\  \displaystyle \large{ \frac{5 \pi}{4}  + 4\pi =  \frac{21 \pi}{4} } \\  \displaystyle \large{ \frac{5 \pi}{4}  + 6\pi =  \frac{29 \pi}{4} }

We cannot use any further k-values (or k cannot be 4 or higher) because it'd be +8π and not in the interval.

<u>F</u><u>o</u><u>r</u><u> </u><u>Q</u><u>4</u>

\displaystyle \large{ \frac{ 7 \pi}{4}  + 2 \pi =  \frac{15 \pi}{4} } \\  \displaystyle \large{ \frac{ 7 \pi}{4}  + 4 \pi =  \frac{23\pi}{4} } \\  \displaystyle \large{ \frac{ 7 \pi}{4}  + 6 \pi =  \frac{31 \pi}{4} }

Therefore:-

\displaystyle \large{4 \theta =  \frac{5 \pi}{4} , \frac{7 \pi}{4} , \frac{13\pi}{4} , \frac{21\pi}{4} , \frac{29\pi}{4}, \frac{15 \pi}{4} , \frac{23\pi}{4} , \frac{31\pi}{4}  }

Then we divide all these values by 4.

\displaystyle \large \boxed{\theta =  \frac{5 \pi}{16} , \frac{7 \pi}{16} , \frac{13\pi}{16} , \frac{21\pi}{16} , \frac{29\pi}{16}, \frac{15 \pi}{16} , \frac{23\pi}{16} , \frac{31\pi}{16}  }

Let me know if you have any questions!

3 0
3 years ago
what is the solution to set to the inequality (4x-3)(2x-1)_&gt;0 a) {x|x&lt;_3 or x_&gt;1} b) {x|x&lt;_2 or x_&gt; 4/3} c) {x|x&
Vinil7 [7]
       (4x - 3)(2x - 1) ≥ 0
4x - 3 ≥ 0    or    2x - 1 ≤ 0
    + 3  + 3                + 1 + 1
     4x ≥ 3                 2x ≤ 1
      4     4                  2    2
       x ≥ ³/₄                  x ≤ ¹/₂
        x ∈ [-∞, ¹/₂] ∧ [³/₄, ∞]

The answer is C.
6 0
3 years ago
Read 2 more answers
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