3.4×10⁻¹⁴ × 1.8×10²⁸
= (3.4 x 1.8) x 10⁽⁻¹⁴⁺²⁸⁾
= 6.12 x 10¹⁴
Answer is A.
Hope it helps!
A pythagoras triplet is a set of three numbers ... not just any three numbers,
but a set where the (square of one of them) is the sum of the (squares of the
other two).
If they're related in that way, then they can be the lengths of the sides of a
right triangle.
If they're not, then they can't.
Take the augmented matrix,
![\left[\begin{array}{ccc|c}2&1&-3&-20\\1&2&1&-3\\1&-1&5&19\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D2%261%26-3%26-20%5C%5C1%262%261%26-3%5C%5C1%26-1%265%2619%5Cend%7Barray%7D%5Cright%5D)
Swap the row 1 and row 2:
![\left[\begin{array}{ccc|c}1&2&1&-3\\2&1&-3&-20\\1&-1&5&19\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%262%261%26-3%5C%5C2%261%26-3%26-20%5C%5C1%26-1%265%2619%5Cend%7Barray%7D%5Cright%5D)
Add -2(row 1) to row 2, and -1(row 1) to row 3:
![\left[\begin{array}{ccc|c}1&2&1&-3\\0&-3&-5&-14\\0&-3&4&22\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%262%261%26-3%5C%5C0%26-3%26-5%26-14%5C%5C0%26-3%264%2622%5Cend%7Barray%7D%5Cright%5D)
Add -1(row 2) to row 3:
![\left[\begin{array}{ccc|c}1&2&1&-3\\0&-3&-5&-14\\0&0&9&36\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%262%261%26-3%5C%5C0%26-3%26-5%26-14%5C%5C0%260%269%2636%5Cend%7Barray%7D%5Cright%5D)
Multiply through row 3 by 1/9:
![\left[\begin{array}{ccc|c}1&2&1&-3\\0&-3&-5&-14\\0&0&1&4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%262%261%26-3%5C%5C0%26-3%26-5%26-14%5C%5C0%260%261%264%5Cend%7Barray%7D%5Cright%5D)
Add 5(row 3) to row 2:
![\left[\begin{array}{ccc|c}1&2&1&-3\\0&-3&0&6\\0&0&1&4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%262%261%26-3%5C%5C0%26-3%260%266%5C%5C0%260%261%264%5Cend%7Barray%7D%5Cright%5D)
Multiply through row 2 by -1/3:
![\left[\begin{array}{ccc|c}1&2&1&-3\\0&1&0&-2\\0&0&1&4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%262%261%26-3%5C%5C0%261%260%26-2%5C%5C0%260%261%264%5Cend%7Barray%7D%5Cright%5D)
Add -2(row 2) and -1(row 3) to row 1:
![\left[\begin{array}{ccc|c}1&0&0&-3\\0&1&0&-2\\0&0&1&4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%260%26-3%5C%5C0%261%260%26-2%5C%5C0%260%261%264%5Cend%7Barray%7D%5Cright%5D)
So we have 
.
All the steps were correct except the final statement. The
mistake was in Line 6.
Line 6 triangle ABC is congruent to triangle EFD by
SAS.
<span>This does not follow. The SAS postulate states
that if two sides and the included angle of one triangle is congruent to two sides
and the included angle of another triangle. The student only proved that one side
of the triangle (AC) is congruent to the side of another triangle (EF) .</span>
