Answer:
A. 1,2,3
Step-by-step explanation:
The only Pythagorean triple that is an arithmetic sequence is 3, 4, 5 and its multiples.
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The triple 1, 2, 3 is not a Pythagorean triple. (It doesn't even form a triangle.)
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If you don't recognize the offered choices as being among the Pythagorean triples you have memorized, you can always check to see if ...
a² +b² = c²
1² +2² = 1 +4 = 5 ≠ 3² . . . . . (1, 2, 3) is not a Pythagorean triple
3² +4² = 9 +16 = 25 = 5²
5² +12² = 25 +144 = 169 = 13²
7² +24² = 49 +576 = 625 = 25²
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A couple of other triples commonly seen are ...
(8, 15, 17), (9, 40, 41)
First step: name the sides according to geometry standards, namely, the sides are named the same lowercase letter as the opposing angle. A revised diagram is shown.
Second step: we need to know the relationships of the trigonometric functions.
cosine(A)=cos(63) = adjacent / hypotenuse = AC/AB .................(1)
sine(A)=sin(63) = opposite / hypotenuse = CB/AB .......................(2)
We're given AB=7, so
using (1)
AC/AB=cos(63)
AC=ABcos(63)=7 cos(63) = 7*0.45399 = 3.17993 = 3.180 (to three dec. figures)
Using (2)
BC/AB=sin(63)
BC=ABsin(63) = 7 sin(63) = 7*0.89101 = 6.237 (to three dec. figures).
Answer:
7(a + 2b)(a² - 2ab + 4b²)
Step-by-step explanation:
Given
7a³ + 56b³ ← factor out 7 from each term
= 7(a³ + 8b³) ← sum of cubes which factors in general as
a³ + b³ = (a + b)(a² - ab + b²)
8b³ = (2b)³ ⇒ b = 2b
a³ + 8b³ = (a + 2b)(a² - 2ab + (2b)²) = (a + 2b)(a² - 2ab + 4b²)
Hence
7a³ + 56b³ = 7(a + 2b)(a² - 2ab + 4b²) ← in factored form