Answer:
a 10 yd 2ft 7 in
b 10 yd 1 ft 2 in
c 99 yd
d. 4840 sq yd 7 sq ft 119 sq in
Step-by-step explanation:
a. 2 ft. 5 in. +
9 yd. 3 ft. 2 in.
---------------------
9 yd 5ft 7 in
but 3 ft = 1yd
9 yd 5ft 7 in
+1yd - 3ft
---------------------------
10 yd 2ft 7 in
b. 4 yd. 8 in
+ 6 yd. 6 in.
----------------------
10 yd 14 in
but 12 in = 1ft
10 yd 14 in
+ 1ft - 12 in
---------------------
10 yd 1 ft 2 in
c. 29 yd. 2 ft. 11 in.
55 yd. 1 ft. 10 in.
+ 13 yd. 1 ft. 3 in.
--------------------------
97 yd 4 ft 24 in
12 in = 1ft
24 in = 2ft
97 yd 4 ft 24 in
+2ft - 24 in
--------------------------------
97 yd 6ft
3 ft = 1yd
6f = 2yd
97 yd 6ft
+2yd - 6ft
---------------------
99 yd
d. 4,839 sq. yd. 8 sq. ft. 139 sq. in.
+ 7 sq. ft. 124 sq. in.
--------------------------------------------------
4839 sq yd 15 sq ft 263 sq in
12*12 = 144 sq in = 1 sq ft
4839 sq yd 15 sq ft 263 sq in
+ 1 sq ft - 144 sq in
--------------------------------------------------
4839 sq yd 16 sq ft 119 sq in
3ft * 3 ft = 9 sq ft = 1 sq yd
4839 sq yd 16 sq ft 119 sq in
+1 sq yd - 9 sq ft
----------------------------------------------
4840 sq yd 7 sq ft 119 sq in
Answer:
2.5
Step-by-step explanation:
Answer: hello your question is poorly written hence I will provide the required matrix
answer :
A = ![\left[\begin{array}{ccc}1&0&1\\0&1&1\\1&-1&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%261%5C%5C0%261%261%5C%5C1%26-1%260%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
Given that the basis of the orthogonal complement have been provided already by you in the question I will have to provide the Matrix
The required matrix
column1 = column 3 - column2
where column 3 and column 2 are the basis of the orthogonal complement of the column space of the Matrix
The instructions are cut off, but if segment PQ bisects angle SQR, then you are correct in saying that angle 1 and angle 2 are congruent to one another.
Answer:
b 15 - 4i sqrt(2)
Step-by-step explanation:
15 - sqrt (-32)
15 - sqrt(-1) sqrt(32)
15 - ±i sqrt(16) sqrt(2)
15 - ±4i sqrt(2)
Taking the principal square root
15 - 4i sqrt( 2)
