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Q: What is the value of y when y 2 e and x 0?
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What is the value of x when 4(x-2)0?

If you mean: 4(x-2) = 0 then 4x-8 = 0 and x = 2


What is the value of 2 x 0 x 1 plus 1?

2 x 0 x 1 + 1 = 0 + 1 = 1


What is the value of e to the power of (2x) when e to the power of x equals 2?

The power law of indices says: (x^a)^b = x^(ab) = x^(ba) = (x^b)^a → e^(2x) = (e^x)² but e^x = 2 → e^(2x) = (e^x)² = 2² = 4


How do you solve absolute value equations with x on both sides?

Possible example: 2|x| = x2 - 8 For x>=0, x2 - 2x - 8 = 0 (x-4)(x+2)=0 so x=4 (can't be -2 since x>=0) For x<0, x2 + 2x - 8 = 0 (x-2)(x+4)=0 so x=-4 (can't be 2 since x<0)


What is the absolute value x-3-6 equals 2?

If: |x|-3-6 = 2 Then: |x| = 2+3+6 = 11 If x > 0, then x = 11 If x < 0, then x = -11


What is the value of y if x equals 0?

Then, y can be any value such that x = 0! If that equation doesn't contain y values, then this means that any y value work for the equation! For instance, if y = 1, then x = 0. If y = 2, then x = 0 and so on.


What does V 3.14 x 7.0 2 20.0 equal?

0.e+0 4.e+1


10010011 base 2 to decimal?

In binary, each place value column is twice the value of the one on its right. Thus 10010011 {in base 2} = 1 x 128 + 0 x 64 + 0 x 32 + 1 x 16 + 0 x 8 + 0 x 4 + 1 x 2 + 1 x 1 = 128 + 16 + 2 + 1 = 147


What is the vertex for the parabola y equals x squared plus 4x plus 5?

The vertex of a parabola is the minimum or maximum value of the parabola. To find the maximum/minimum of a parabola complete the square: x² + 4x + 5 = x² + 4x + 4 - 4 + 5 = (x² + 4x + 4) + (-4 + 5) = (x + 2)² + 1 As (x + 2)² is greater than or equal to 0, the minimum value (vertex) occurs when this is zero, ie (x + 2)² = 0 → x + 2 = 0 → x = -2 As (x + 2)² = 0, the minimum value is 0 + 1 = 1. Thus the vertex of the parabola is at (-2, 1).


How you write 22000000 in expanded form?

To write 22000000 in expanded form, you would break it down by place value. It can be expressed as 2 * 10,000,000 + 2 * 1,000,000 + 2 * 100,000 + 2 * 10,000 + 2 * 1,000.


What is the derivative of e to the power of 2lnx?

Your expression simplifies to just x^2 {with the restriction that x > 0}. The derivative of x^2 is 2*x