In
order to solve for a nth term in an arithmetic sequence, we use the formula
written as:<span>
an = a1 + (n-1)d
where an is the nth term, a1 is the first value
in the sequence, n is the term position and d is the common difference.
</span><span>THIRD
</span><span>A3=4+(3-1)(-5)
A3 = -6
A(3)=-2(3-1)(-5)
A3 = 20
</span><span>
FIFTH
</span>A5=4+(5-1)(-5)
A5 = -16
A(5)=-2(5-1)(-5)
A5 = 40<span>
TENTH
</span>A10=4+(10-1)(-5)
A10 = -41
A(10)=-2(10-1)(-5)
A10 = 90
206 X .4 = 82.4
That's the answer.
Hi.. U should do like this
4/5 mile in 50 min
X mile in 60 min
X=5/4 mile
Answer:
8
Step-by-step explanation:
8x
40=8*5
The greatest common factor of 8x and 40 is 8
Step-by-step explanation:
The area would be 9 times compared to the area of the original square. To test this, you can let the side of the original square be equal 1. By tripling this side, the side becomes three. Utilizing the area of a square formula, A= s^2, the area of the original square would be 1 after substituting 1 for s. Then, you do the same for the area of the tripled square. With the substitution, the area of the tripled square would be 9. This result displays the area of the tripled square being 9 times as large as the area of the original square. This pattern can be used for other measurements of the square such as:
let s = 2, Original Area= 2^2 = 4 Tripled Area= (2(3))^2 = 6^2= 36. 36/4 = 9
let s = 3, Original Area = 3^2 = 9 Tripled Area - (3(3))^2 = 9^2 =81. 81/9 = 9
let s = 4, Original Area = 4^2 = 16 Tripled Area - (4(3))^2 = 12^2 = 144. 144/16 = 9
let s = 5, Original Area = 5^2 = 25 Tripled Area - (5(3))^2 = 15^2 = 225. 225/25 = 9
let s = 6, Original Area = 6^2 = 36 Tripled Area - (6(3))^2 = 18^2 = 324. 324/36 = 9
let s = 7, Original Area = 7^2 = 49 Tripled Area - (7(3))^2 = 21^2 = 2,401. 2,401/49 = 9
You can continue to increase the length of the square and follow this pattern and it will be consistent.
