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Square root 24x4 divided by square root 3x?
Sqrt(24x4)/Sqrt(3x) = Sqrt(24x4/3x) = Sqrt(8x3) = Sqrt(4x2*2x) = 2x*sqrt(2x)
Solution set for 4x2-6 equals 2x2?
Reading this as: 4x2-6=2x2 This implies that: -6=-2x2 3=x2 x=sqrt(3) and x=-sqrt(3) So x can equal plus or minus the square root of 3
What is the absolute value of 7 5i?
The absolute value of 7± 5i = sqrt(72 + 52) = sqrt(49 + 25) = sqrt(74) = 8.602, approx.
What is the exact value of sine 15?
(sqrt(6)-sqrt(2))/4
What is the exact value of cosine -15?
(1+sqrt(3))/(2 sqrt(2))
Square root 24x4 divided by square root 3x?
Sqrt(24x4)/Sqrt(3x) = Sqrt(24x4/3x) = Sqrt(8x3) = Sqrt(4x2*2x) = 2x*sqrt(2x)
Sqrt 24x 4th divided by sqrt 3x equals?
sqrt(24x4) / sqrt(3x) = sqrt(24x4/3x) = sqrt(8x3) = sqrt(2x times 4x2) = 2x sqrt(2x) or (2x)1.5
Y equals 4x2 plus 6x - 1?
Y = 4x2 + 6x - 1Use the pq-formula:x = -4/2 +- sqrt ((4/2)2 -(-1))x = -2 +- sqrt((16/4) +1)x = -2 +- sqrt(4 +1)x1 = -2 + sqrt(5)x2 = -2 - sqrt(5)
Factor the following polynomial 4x2 - 169?
sqrt(4)=2 sqrt(169)=13 (2x+13)(2x-13)
Solution set for 4x2-6 equals 2x2?
Reading this as: 4x2-6=2x2 This implies that: -6=-2x2 3=x2 x=sqrt(3) and x=-sqrt(3) So x can equal plus or minus the square root of 3
What is the absolute value of 7i?
The absolute value is sqrt(72 + 12) = sqrt(49 + 1) = sqrt(50) or 5*sqrt(2) = 7.071 approx.
How do you write 4x2 equals 12 in standard form?
4x^2 = 12 Divide through by '4' x^2 = 3 Square root both sides x = +/-sqrt(3)
What is the absolute value of the complex number 5 - 4i?
1
Could you help me to 4x2-4x plus 1 equals 9?
Probably a quadratic equation problem.4X2 - 4X + 1 = 94X2 - 4X - 8 = 0factor out a 4 looks like a dead end, so....X = - b (+/-) sqrt(b2 - 4ac)/2aa = 4b = - 4c = - 8X = - (- 4) (+/-) sqrt[(- 4)2 -4(4)(- 8)]/2(4)X = 4 (+/-) sqrt(16 + 128)/8X = 4 (+/-) sqrt(144)/8X = [4 (+/-) 12]/8================X = 2X = - 1
What is the absolute value of -7-5i?
-2
What is the absolute value of 7 5i?
The absolute value of 7± 5i = sqrt(72 + 52) = sqrt(49 + 25) = sqrt(74) = 8.602, approx.
What is value of square root of negative iota?
-i/sqrt(2) -i/sqrt(2)
