Answer:
97.98
Step-by-step explanation:
The area of the parallelogram PQR is the magnitude of the cross product of any two adjacent sides. Using PQ and PS as the adjacent sides;
Area of the parallelogram = |PQ×PS|
PQ = Q-P and PS = S-P
Given P(0,0,0), Q(4,-5,3), R(4,-7,1), S(8,-12,4)
PQ = (4,-5,3) - (0,0,0)
PQ = (4,-5,3)
Also, PS = S-P
PS = (8,-12,4)-(0,0,0)
PS = (8,-12,4)
Taking the cross product of both vectors i.e PQ×PS
(4,5,-3)×(8,-12,4)
PQ×PS = (20-36)i - (16-(-24))j + (-48-40)k
PQ×PS = -16i - 40j -88k
|PQ×PS| = √(-16)²+(-40)²+(-88)²
|PQ×PS| = √256+1600+7744
|PQ×PS| = √9600
|PQ×PS| ≈ 97.98
Hence the area of the parallelogram is 97.98
Slope intercept form: y=Mx+b (where m is the slope and b is the y intercept)
So just put it into the equation
Answer: y= -8x + 4
<u>Answer-</u>
<em>D. The function has one real zero and two nonreal zeros. The graph of the function intersects the x-axis at exactly one location.</em>
<u>Solution-</u>
The given polynomial is,
The zeros of the polynomials are,
Therefore, this function has only one real zero i.e 1 and two nonreal zeros i.e ±√6i . The graph of the function intersects the x-axis at exactly one location i.e at x = 1
Answer:
D. RWS and SWT
Step-by-step explanation:
Adjacent angles have a common side and a common vertex but they don't overlaps each other
The two angles that has these requirements are :
RWS and SWT
Answer:
(2x+2)
Step-by-step explanation:
