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iren [92.7K]
3 years ago
14

Find an equation for the nth term of the arithmetic sequence. a19 = -58, a21 = -164

Mathematics
1 answer:
Elodia [21]3 years ago
8 0
So the formula would be -1856 +98[n-1]
You might be interested in
Angle 0 lies in the second quadrant and sin 0 = 3/5 cos= coy =
Elza [17]
Consider the right triangle ABC with legs AB=4, AC=3 and hypotenuse BC=5. Angle B has
 \sin B=\frac{opposite}{hypotenuse} = \frac{3}{5} and
 \cos B = \frac{adjacent}{hypotenuse} = \frac{4}{5}.

Since O lies in second quadrant \cos O\ \textless \ 0 and 
\cos O= -\cos B =- \frac{4}{5}.
Answer: \cos O=- \frac{4}{5}. 


3 0
3 years ago
PLEASE HELP!! If the endpoints of have the coordinates A(9, 8) and B(-1, -2), what is the midpoint of ?
dimaraw [331]

M refers to midpoint.

A(9, 8) \\B(-1, -2) \\M(\frac{x_1-x_2}{2}, \frac{y_1-y_2}{2}) \\M(\frac{9-(-1)}{2}, \frac{8-(-2)}{2}) \\M(\frac{10}{2}, \frac{10}{2})\Longrightarrow\boxed{M(5, 5)}

Hope this helps.

r3t40

6 0
3 years ago
Simplify (6^5/7^3)^2
o-na [289]
Simplify the following:
(6^5/7^3)^2

7^3 = 7×7^2:
(6^5/(7×7^2))^2

7^2 = 49:
(6^5/(7×49))^2

7×49 = 343:
(6^5/343)^2

6^5 = 6×6^4 = 6 (6^2)^2:
((6 (6^2)^2)/343)^2

6^2 = 36:
((6×36^2)/343)^2

 | | 3 | 6
× | | 3 | 6
 | 2 | 1 | 6
1 | 0 | 8 | 0
1 | 2 | 9 | 6:
((6×1296)/343)^2

6×1296 = 7776:
(7776/343)^2

(7776/343)^2 = 7776^2/343^2:
7776^2/343^2

 | | | | 7 | 7 | 7 | 6
× | | | | 7 | 7 | 7 | 6
 | | | 4 | 6 | 6 | 5 | 6
 | | 5 | 4 | 4 | 3 | 2 | 0
 | 5 | 4 | 4 | 3 | 2 | 0 | 0
5 | 4 | 4 | 3 | 2 | 0 | 0 | 0
6 | 0 | 4 | 6 | 6 | 1 | 7 | 6:
60466176/343^2

 | | | 3 | 4 | 3
× | | | 3 | 4 | 3
 | | 1 | 0 | 2 | 9
 | 1 | 3 | 7 | 2 | 0
1 | 0 | 2 | 9 | 0 | 0
1 | 1 | 7 | 6 | 4 | 9:

Answer: 60466176/117649
6 0
2 years ago
Solve and Simplify: 8 - 6_3/9
Mars2501 [29]

Answer:  

Exact form: 3/5

Decimal form: 1.6

Mixed number form: 1_ 2/3

6 0
2 years ago
Pls answer the question fast
Rasek [7]

Answer:

d

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
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