In ΔOPQ, o = 9.2 cm, p = 2.4 cm and ∠Q=37°. Find the length of q, to the nearest 10th of a centimeter.
1 answer:
The length of q, to the nearest 10th of a centimeter is 7.6 cm.
Given in question,
In ΔOPQ,
o = 9.2 cm
p = 2.4 cm
∠Q = 37°
Cosine formula ⇒ cos θ =
Putting the values in equation,
cos 37 =
0.799 =
0.799*44.16 = 90.4 -
32.28 = 90.4 -
= 90.4 - 32.28
= 58.12
q = 7.6 cm (to nearest 10th)
Hence, length of q is 7.6 cm.
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Answer:
The remainder would be 1 or 3
Step-by-step explanation:
Consider the 5 consecutive positive integers are,
x, x + 1, x + 2, x + 3, x + 4,
Since, the average of these 5 numbers =
= x + 2
If x + 2 = odd,
⇒ x = odd - 2 = odd - even = odd
⇒ x + 4 = odd + even = odd
∵ an odd number is represented by '2n + 1'
Where, n = 0, 1, 2, 3, ........
Now, 2n + 1 = 1( mod 4) if n = even
While, 2n + 1 = 3( mod 4) if n = odd,
Hence, when the largest of the five integers is divided by 4 remainder would be 1 or 3.