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What is 3x4 plus 5x3 plus x2 - 5 divided by x 2?
3x4 plus 5x3 plus x2 - 5 divided by x 2 =[(3x4) + (5x3) + (x2 - 5)]/x2 =(12 + 15 + x2 -5)/x2 =(27 - 5 + x2)/x2 =(22 + x2)/x2
What is x to the 2nd power -3x plus2 times x minus 2?
Hmm - you need to phrase your question a little better. Which of these are you looking for? x2 - 3x + 2x - 2 = x2 - x - 2 = (x - 2)(x + 1) x2 - (3x + 2)(x - 2) = x2 - 3x2 + 6x - 2x + 4 = -2x2 + 4x + 4 = -2(x2 - 2x - 2) x2 - 3x + 2(x - 2) = x2 + x - 4
M is the midpoint of pq verify that this is the midpoint by using the distance formula to show tha pm equals mq?
Let P(x1, y1), Q(x2, y2), and M(x3, y3).If M is the midpoint of PQ, then,(x3, y3) = [(x1 + x2)/2, (y1 + y2)/2]We need to verify that,√[[(x1 + x2)/2 - x1]^2 + [(y1 + y2)/2 - y1]^2] = √[[x2 - (x1 + x2)/2]^2 + [y2 - (y1 + y2)/2]^2]]Let's work separately in both sides. Left side:√[[(x1 + x2)/2 - x1]^2 + [(y1 + y2)/2 - y1]^2]= √[[(x1/2 + x2/2)]^2 - (2)(x1)[(x1/2 + x2/2)) + x1^2] + [(y1/2 + y2/2)]^2 - (2)(y1)[(y1/2 + y2/2)] + y1^2]]= √[[(x1)^2]/4 + [(x1)(x2)]/2 + [(x2)^2]/4 - (x1)^2 - (x1)(x2) + (x1)^2 +[(y1)^2]/4 + [(y1)(y2)]/2 + [(y2)^2]/4 - (y1)^2 - (y1)(y2) + (y1)^2]]= √[[(x1)^2]/4 - [(x1)(x2)]/2 + [(x2)^2]/4 + [(y1)^2]/4 - [(y1)(y2)]/2 + [(y2)^2]/4]]Right side:√[[x2 - (x1 + x2)/2]^2 + [y2 - (y1 + y2)/2]^2]]= √[[(x2)^2 - (2)(x2)[(x1/2 + x2/2)] + [(x1/2 + x2/2)]^2 + [(y2)^2 - (2)(y2)[(y1/2 + y2/2)] + [(y1/2 + y2/2)]^2]]= √[[(x2)^2 - (x1)(x2) - (x2)^2 + [(x1)^2]/4 + [(x1)(x2)]/2 + [(x2)^2]/4 + (y2)^2 - (y1)[(y2) - (y2)^2 + [(y1)^2]/4) + [(y1)(y2)]/2 + [(y2)^2]/4]]= √[[(x1)^2]/4 - [(x1)(x2)]/2 + [(x2)^2]/4 + [(y1)^2]/4 - [(y1)(y2)]/2 + [(y2)^2]/4]]Since the left and right sides are equals, the identity is true. Thus, the length of PM equals the length of MQ. As the result, M is the midpoint of PQ
X2 -2 equals 0?
If "X2-2" refers to X * 2 - 2, the answer is X = 1; If "X2-2" refers to "X squared" - 2, the answer is X = square root of 2 (√2)
Can someone solve for x. 2x2 equals x2-2?
2x2 = x2 - 2 Subtract x2 from both sides: x2 = -2 Take square roots: x = + or - i*sqrt(2) where i is the imaginary sqrt of -1.
What is the greatest common factor of x9 and x2?
The GCF is x2.
Sum of series 1 plus x2 plus x4.....plus xn?
(xn+2-1)/(x2-1)ExplanationLet Y=1+x2+x4+...+xn. Now notice that:Y=1+x2+x4+...+xn=x2(1+x2+x4+...+xn-2)+1Y+xn+2=x2(1+x2+x4+...+xn-2+xn)+1Y+xn+2=x2*Y+1Y+xn+2-x2*Y=1Y-x2*Y=1-xn+2Y(1-x2)=1-xn+2Y=(1-xn+2)/(1-x2)=(xn+2-1)/(x2-1)
What is the anti derivative of x divided by x squared minus 1?
The antiderivative of x/(x2-1) is ln(x2-1)/2. Proof: (ln(x2-1)/2)' = (1/(x2-1))*(x2-1)'/2=1/(x2-1)*(2x/2)=x/(x2-1).
What is 4x-x2?
4x-x2 = 2
How do you factor x2 6x-2?
x2 + 6x - 2 can not be factored
What is 3x4 plus 5x3 plus x2 - 5 divided by x 2?
3x4 plus 5x3 plus x2 - 5 divided by x 2 =[(3x4) + (5x3) + (x2 - 5)]/x2 =(12 + 15 + x2 -5)/x2 =(27 - 5 + x2)/x2 =(22 + x2)/x2
What is x to the 2nd power -3x plus2 times x minus 2?
Hmm - you need to phrase your question a little better. Which of these are you looking for? x2 - 3x + 2x - 2 = x2 - x - 2 = (x - 2)(x + 1) x2 - (3x + 2)(x - 2) = x2 - 3x2 + 6x - 2x + 4 = -2x2 + 4x + 4 = -2(x2 - 2x - 2) x2 - 3x + 2(x - 2) = x2 + x - 4
M is the midpoint of pq verify that this is the midpoint by using the distance formula to show tha pm equals mq?
Let P(x1, y1), Q(x2, y2), and M(x3, y3).If M is the midpoint of PQ, then,(x3, y3) = [(x1 + x2)/2, (y1 + y2)/2]We need to verify that,√[[(x1 + x2)/2 - x1]^2 + [(y1 + y2)/2 - y1]^2] = √[[x2 - (x1 + x2)/2]^2 + [y2 - (y1 + y2)/2]^2]]Let's work separately in both sides. Left side:√[[(x1 + x2)/2 - x1]^2 + [(y1 + y2)/2 - y1]^2]= √[[(x1/2 + x2/2)]^2 - (2)(x1)[(x1/2 + x2/2)) + x1^2] + [(y1/2 + y2/2)]^2 - (2)(y1)[(y1/2 + y2/2)] + y1^2]]= √[[(x1)^2]/4 + [(x1)(x2)]/2 + [(x2)^2]/4 - (x1)^2 - (x1)(x2) + (x1)^2 +[(y1)^2]/4 + [(y1)(y2)]/2 + [(y2)^2]/4 - (y1)^2 - (y1)(y2) + (y1)^2]]= √[[(x1)^2]/4 - [(x1)(x2)]/2 + [(x2)^2]/4 + [(y1)^2]/4 - [(y1)(y2)]/2 + [(y2)^2]/4]]Right side:√[[x2 - (x1 + x2)/2]^2 + [y2 - (y1 + y2)/2]^2]]= √[[(x2)^2 - (2)(x2)[(x1/2 + x2/2)] + [(x1/2 + x2/2)]^2 + [(y2)^2 - (2)(y2)[(y1/2 + y2/2)] + [(y1/2 + y2/2)]^2]]= √[[(x2)^2 - (x1)(x2) - (x2)^2 + [(x1)^2]/4 + [(x1)(x2)]/2 + [(x2)^2]/4 + (y2)^2 - (y1)[(y2) - (y2)^2 + [(y1)^2]/4) + [(y1)(y2)]/2 + [(y2)^2]/4]]= √[[(x1)^2]/4 - [(x1)(x2)]/2 + [(x2)^2]/4 + [(y1)^2]/4 - [(y1)(y2)]/2 + [(y2)^2]/4]]Since the left and right sides are equals, the identity is true. Thus, the length of PM equals the length of MQ. As the result, M is the midpoint of PQ
If the distance between the point 5-3 and x 1 is 5 units find x?
Final solution, x = 2 ----- WORK ----- (5, -3) (X2, 1) (X1, Y1) (X2, Y2) D = SQRT( [X1 - X2]2 + [Y1 - Y2]2 ) 5 = SQRT( [5 - X2]2 + [-3 - 1]2 ) 5 = SQRT( [5 - X2]2 + (-4)2 ) 5 = SQRT( [5 - X2]2 + 16 ) Since The square root of 25 is 5, we know that -> [5 - X2]2 + 16 = 25 Subtract 16 from both sides, [5 - X2]2 = 9 Which means that 5-X2 = 3 Because 32 is 9 5 - X2 = 3 subtract 5 from both sides -X2 = -2 multiply both sides by -1 X2 = 2 There you go... X2 = 2 (5, -3) (2, 5)
Factor x2-2 plus 4?
x2 - 2 + 4 = x2 + 2This term can not be factored any further
X2 -2 equals 0?
If "X2-2" refers to X * 2 - 2, the answer is X = 1; If "X2-2" refers to "X squared" - 2, the answer is X = square root of 2 (√2)
Can someone solve for x. 2x2 equals x2-2?
2x2 = x2 - 2 Subtract x2 from both sides: x2 = -2 Take square roots: x = + or - i*sqrt(2) where i is the imaginary sqrt of -1.
