Answer:
1.34
Step-by-step explanation:
Used a Calculator
The augmented matrix is
![\left[\begin{array}{ccccc|c}1&-2&3&2&1&10\\2&-4&8&3&10&7\\3&-6&10&6&5&27\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7Cc%7D1%26-2%263%262%261%2610%5C%5C2%26-4%268%263%2610%267%5C%5C3%26-6%2610%266%265%2627%5Cend%7Barray%7D%5Cright%5D)
Add -2(row 1) to row 2, and add -3(row 1) to row 3:
![\left[\begin{array}{ccccc|c}1&-2&3&2&1&10\\0&0&1&-1&8&-13\\0&0&1&0&2&-3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7Cc%7D1%26-2%263%262%261%2610%5C%5C0%260%261%26-1%268%26-13%5C%5C0%260%261%260%262%26-3%5Cend%7Barray%7D%5Cright%5D)
Add -1(row 2) to row 3:
![\left[\begin{array}{ccccc|c}1&-2&3&2&1&10\\0&0&1&-1&8&-13\\0&0&0&1&-6&10\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7Cc%7D1%26-2%263%262%261%2610%5C%5C0%260%261%26-1%268%26-13%5C%5C0%260%260%261%26-6%2610%5Cend%7Barray%7D%5Cright%5D)
Add row 3 to row 2:
![\left[\begin{array}{ccccc|c}1&-2&3&2&1&10\\0&0&1&0&2&-3\\0&0&0&1&-6&10\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7Cc%7D1%26-2%263%262%261%2610%5C%5C0%260%261%260%262%26-3%5C%5C0%260%260%261%26-6%2610%5Cend%7Barray%7D%5Cright%5D)
Add -3(row 2) to row 1, and add -2(row 3) to row 1:
![\left[\begin{array}{ccccc|c}1&-2&0&0&7&-1\\0&0&1&0&2&-3\\0&0&0&1&-6&10\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7Cc%7D1%26-2%260%260%267%26-1%5C%5C0%260%261%260%262%26-3%5C%5C0%260%260%261%26-6%2610%5Cend%7Barray%7D%5Cright%5D)
<h3>
Answer:</h3>
3 ft
<h3>
Step-by-step explanation:</h3>
The statue's height is 1.5 times the length of its shadow, so we expect the same relationship for the globe.
... 1.5 × 2 ft = 3 ft
_____
<em>Comment on the problem</em>
As a practical matter, with the sun high enough in the sky to cast a shadow shorter than the object's height, it will be quite difficult to measure the length of the shadow of the point at the top of the globe. The shadow of other parts of the globe will interfere.
X = 4y - 18
-3(4y - 18) + 12y =54
-12y + 54 + 12y = 54
54 = 54
c
<u>Given </u><u>:</u><u>-</u><u> </u>
- An extension ladder with 3 five-foot extensions is leaned against a wall with the base of the ladder 5 feet from the base of the wall.
<u>To </u><u>find</u><u> </u><u>:</u><u>-</u>
- How high on the wall does the top of the ladder sit?
<u>Solution</u><u> </u><u>:</u><u>-</u><u> </u>
Using Pythagoras Theorem ,
- 5² = 3² + h²
- 25 = 9 + h ²
- h² = 25 - 9
- h ²= 16
- h =√16
- h = ±4
- h = 4
