Answer:
‹A = 141°
‹B = 28°
‹C = 11°
*These angles are rounded to the nearest whole*
Step-by-step explanation:
Because you are only given sides you can find an individual angle measures with the inverse law of cosines.
Remember the side opposite of the angle corresponds to that angle.
a² = b² + c² - 2bc cos(A) →
A = cos⁻¹ (a² - b² - c² / -2bc)
b² = a² + c² - 2ac cos(B) →
B = cos⁻¹ (b² - a² - c² / -2ac)
c² = a² + b² - 2ab cos(C) →
C = cos⁻¹ (c² - a² - b² / -2ab)
Solve for x over the real numbers:
x^3 (x^2 - 4) = 0
Split into two equations:
x^3 = 0 or x^2 - 4 = 0
Take cube roots of both sides:
x = 0 or x^2 - 4 = 0
Add 4 to both sides:
x = 0 or x^2 = 4
Take the square root of both sides:
Answer: x = 0 or x = 2 or x = -2
Answer:
Jada's book was overdue by 1 more day than Juan's book.
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Answer:
Check the ecplanation
Step-by-step explanation:
A set of three vectors in 
represents a matrix of 3 column vectors, and each vector containing 4 entries (that is, a matrix of 4 rows, and 3 columns).
Let A be that 4x 3 matrix. The columns of A span 
. if and only if A has a pivot position in each row. So, there are at most 3 pivot positions in the matrix A, but the number of rows is 4, therefore, there exist at least one row not having a pivot position. If A does not have a pivot position in at least one row, then the columns of A do not span . It implies that the set of 3 vectors of A does not span all of .
In general, the set of n vectors in represents a matrix of in rows, and n columns (an in x matrix). So, there are at most n pivot positions in the matrix A, but n is less than the number of rows. In therefore, there exist at least one row that does not contain a pivot position.
And, hence the set of n vectors of A does not span all of . for n < m
