<span>We know that for two vectors u and v we have that the cross product is equal to :
(u2v3 - u3v2) i + (u3v1 - u1v3) j + (u1v2 - u2v1) k
Then we have that m x n : ( (2 *-4) - (-5 * -2) ) i + ( (-5*6) - (3 *-4) )j + ( ( 3*-2) - (2*6))k
which becomes ( -8 - 10) i + ( -30 + 12) j + ( -6 - 12) k
Thus the vector product of m and n is -18 i - 18j -18k</span>
y=x^2-2x-15 (1) y=8x-40 (2)
8x-40=x^2-2x-15
X^2+10x-25=0
(x-5)^2
×=5 y=0
-10w - 400 = o
w= weeks
o = new outstanding balance
After you have plotted the graph of the polynomial, the x-intercepts of the given polynomial will be:
(-3,0),(-1,0) and (4,0)
therefore the interval will be:
(-∞,-3)-below, (-3,-1)-above,(-1,4)-below,(4,∞)-above
The answer is ]
We need to assign a value for x to check the possible values of y.
1st inequality: y < -0.75x
X = - 1 ; y < -0.75(-1) ; y < 0.75 possible coordinate (-1,0.75) LOCATED AT THE 2ND QUADRANT
X = 0 ; y < -0.75(0) ; y < 0 possible coordinate (0,0) ORIGIN
X = 1 ; y < -0.75(1) ; y < -0.75 possible coordinate (1,-0.75) LOCATED AT THE 4TH QUADRANT
2nd inequality: y < 3x -2
X = -1 ; y < 3(-1) – 2 ; y < -5 possible coordinate (-1,-5) LOCATED AT THE 4TH QUADRANT
X = 0 ; y < 3(0) – 2 ; y < -2 possible coordinate (0,-2) LOCATED AT THE 4TH QUADRANT
X = 1 ; y < 3(1) – 2 ; y <<span> 1 possible coordinate (1,1) LOCATED AT THE 1ST QUADRANT
The actual solution to the system lies on the 4TH QUADRANT.
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