Answer:
The answer to your question is 2i
Step-by-step explanation:
![\sqrt[4]{-16}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B-16%7D)
- Get the prime factors of -16
- 16 2
- 8 2
- 4 2
- 2 2
1
16 = 2⁴
- Express the -16 as a power
![\sqrt[4]{-2^{4}} or \sqrt[4]{2^{4}i^{2}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B-2%5E%7B4%7D%7D%20%20or%20%20%20%5Csqrt%5B4%5D%7B2%5E%7B4%7Di%5E%7B2%7D%7D)
Remember that -1 = i²
- Get the fourth root of -16
2i
The answer is A because of the bar graph
Answer:
The answer is given below
Step-by-step explanation:
A) Integers are whole numbers (without fraction) that are either positive or negative. If T={x:X is an integer between 1 and 4}, therefore the elements in set T = {2, 3}
B) Since the elements in set T = {2, 3}, then the number of elements in set T = 2
C) The subsets of set T are {}, {2}, {3} and {2,3}
D) Proper subset of set T are subsets of T that is not equal to T. The proper subsets of T are {2} and {3}
An improper subset of set T contains all the element of set T and a null element. The improper subset of set T are {2,3} and {}
Answer:
Step-by-step explanation:
57-4=53
53+9=62
Answer:
Answer:
The cross product is 47 i - 61 j + 17 k.
Step-by-step explanation:
Given that-
a = < -1, -3, -8 > , b = < 7, 4, -5 >
The cross product is
a × b =\begin{gathered}\left[\begin{array}{ccc}i&j&k\\-1&-3&-8\\7&4&-5\end{array}\right]\end{gathered}⎣⎢⎡i−17j−34k−8−5⎦⎥⎤
= i ( - 3 × -5 - 4 × -8 ) - j ( -1 × -5 - (-8) × 7 ) + k ( -1 × 4 - 7 × -3 )
= i ( 15 + 32 ) -j ×( 5 + 56 ) + k ( -4 + 21 )
= i ( 47) - j (61 ) + k (17)
= 47 i -61 j + 17 k
As we know that if two vectors A & B are perpendicular then
A . B = 0
So here only the given vector b = < 7,4, -5> is perpendicular to the resulting vector .
Step-by-step explanation:
47, -61, 71>; yes
