Answer:
v=9i+3j
Step-by-step explanation:
The given vector, v has initial point at P1 = (−5, −2) and terminal point at P2 = (4, 1).
The vector v is found by subtraction the initial point from the terminal point.
v=<4,1>-<-5,-2>
v=<4--5,1--2>
v=<9,3>
We write v as multiples of the basis vectors to obtain:
v=9i+3j
Answer:
<h3>C. They are both perfect squares and perfect cubes.</h3>
Step-by-step explanation:
Perfect squares are numbers that their square root can be found easily without any remainder.
Given the following patterns;
1*1 = 1 and 1*1*1 = 1
It can be seen that 1 is 1 perfect square since 1*1 = 1² = 1
Also 1 is perfect cube since 1*1*1 = 1³ = 1 (cube of the value gives 1)
Similarly for the expression;
8*8 = 64
8² = 64 (since the square of 8 gives 64, then 64 is known to be a perfect square)
Also 4*4*4 = 64
i.e 4³ = 64 (This shows that the cube root of 64 is 4 making it a perfect cube since we can get a whole number for the cube root of 64)
The same is applicable for other expressions 729 = 27 × 27, and 9 × 9 × 9, 4,096 = 64 × 64, and 16 × 16 × 16
This values are easily expressed as a constant multiple of a number showing that they are both perfect squares and perfect cubes.
No because 77 is greater than the lowest number
