B<0, an absolute value, |-b|=-b. Why is that so?
If b is negative then you have two minuses in the definition, one inside b, the other one in front of b.|-5| = -(-5) = 5
What is the integral of absolute value of (x-1)?
integral of (x-1) dx =x^2/2 -x =x(x/2 -1) during the interval 0 to 3 the value will be 3(3/2 -1)=3/2 which is option b
Find x: 2I3-5xI = 3I1-2xI (IxI absolute value)?
When solving for an absolute value on both sides, the solution involves 4 possibilities. a) Both absolutes could be positive. 2(3-5x)=3(1-2x) b) Both absolutes could be negative: 2(-(3-5x))=3(-(1-2x)) c) The left side absolute could be negative and the right side absolute could be positive: 2(-(3-5x))=3(1-2x) d) The left side absolute could be positive and the right side absolute could be negative: 2(3-5x)=3(-(1-2x)) Now you solve for each condition to find the corresponding x value. You must be sure to check each answer in the original problem to ensure no extraneous conditions occur. a) 2(3-5x)=3(1-2x) 6-10x=3-6x 3=4x x=3/4 b) 2(-(3-5x))=3(-(1-2x)) 2(5x-3)=3(2x-1) 10x-6=6x-3 4x=3 x=3/4 c) 2(-(3-5x))=3(1-2x) 2(5x-3)=3-6x 10x-6=3-6x 16x=9 x=9/16 d) 2(3-5x)=3(-(1-2x)) 6-10x=3(2x-1) 6-10x=6x-3 9=16x x=9/16 Therefore, the solution is x=3/4 and x=9/16 Hope this helps.
What is the value of the absolute value of 4-3i?
[math] |z| = \sqrt{(x^2 + y^2)} [/math]Substituting the values of ‘x’ and ‘y’ in the formula :[math] |z| = \sqrt{(4^2 + (-3)^2} [/math][math] |z| = \sqrt{ 16 + 9}[/math][math] |z| = \sqrt{25}[/math][math] |z| = \pm 5 [/math]As we know that the absolute value of any number, is its distance from the origin. The distance has to be positive. Thus,[math] |z| = 5 [/math].That is it !!!!With regards.
Integral of an absolute value function?
∫ | (x(^2)-4) | dx a=0,b=7 for the integral Consider (x^2 - 4) = (x-2)(x+2) IN (0,7) this will be +ve if x > 2 otherwise -ve Now |u| = u if u > 0 & -u if u < 0 Therefore ∫ | (x(^2)-4) | dx a=0,b=7 for the integral = { ∫ (x(^2)-4) dx a=2,b=7 for the integral } + { ∫ -(x(^2)-4) dx a=0,b=2 for the integral } = { ∫ (x(^2)-4) dx a=2,b=7 for the integral } + { ∫ (4 - x(^2) dx a=0,b=2 for the integral } = x^3/3 - 4x from 2 to 7 + ( 4x - x^3/3 from 0 to 2 ) Leave the rest to u
How do you convert a compound inequality to an absolute value inequality?
Don't you mean "and"? All real x satisfy at least one of 4x < 16 and 8x > 16 (and some x values satisfy both, which are the "and" solutions). So I assume you want to find an absolute value inequality that is equivalent to 4x < 16 and 8x > 16. 4x < 16 x < 4 8x > 16 x > 2 The compounding of these is 2 < x < 4 You want this in the form |x - a| < b (where b ≥ 0 by definition of absolute value; otherwise, there would be no solution). . Well, let's expand the latter: |x - a| < b -b < x - a < b a - b < x < a + b Therefore, for this example, a - b = 2 a + b = 4 This is easily solved: a = 3, b = 1 Thus, the absolute value inequality is |x - 3| < 1 For a problem like 4x > 16 or 8x < 16, the appropriate absolute value inequality has the form |x - a| > b (where b ≥ 0, by definition of absolute value; otherwise, all real x are solutions) which is equivalent to x - a > b or x - a < -b. Solving these for x, x > a + b or x < a - b For the example problem, we have x > 4 or x < 2, so a - b = 2 and a + b = 4, the solution of which is a = 3, b = 1. So the absolute value inequality form of 4x > 16 or 8x < 16 is |x - 3| > 1
If b<0 and |b|=4b+15, then what is the value of b?
We can rewrite the equation |b|=4b+15 as b=4b+15 and -b=4b+15. This is because an absolute value takes a negative value and makes it positive, ex -6 -> 6. Now we can solve each of the equations:b=4b+15Minus 4b on both sides to isolate the variable-3b=15Divide by -3b=-5-b=4b+15-5b=15b=-3We can check these values by plugging both of them into the equation|b|=4b+15|-3|=4(-3)+153=-12+153=3|b|=4b+15|-5|=4(-5)+155=-20+155=-5This is false which is because the part of the equation that says b<0, in our equation we had a plus b which does not fit that parameter.This means that b=-3
Why does [math]\sqrt{x^2}=|x|[/math] (absolute value) while [math](\sqrt{x})^2=x[/math] ?
So why does[math] \sqrt{x^2} = |x|[/math] but [math]([/math][math]\sqrt{x})^2 = x[/math]?In short, it’s because the square root function always selects the positive root. That’s why there’s the absolute value in the first equation. The absolute value goes away in the second equation because it is squared, which eliminates the sign distinction (e.g. [math](-c)^2 = (-1)^2 c^2 = c^2[/math]).Before jumping into the detailed explanation observe that if [math]x \ge 0[/math] it doesn’t really matter whether we use |x| or x.[math]\sqrt{3^2} = \sqrt{9} = 3[/math][math]\sqrt{3}^2 = 3[/math]But if x < 0 we see the difference.[math]\sqrt{(-3)^2} = \sqrt{9} = 3 = |-3|[/math][math]\sqrt{-3}^2 = (\sqrt{3}i)^2 = -3[/math]Why does it work if [math]x \ge 0[/math] but not if [math]x \lt 0[/math]? Because the square root function selects (somewhat arbitrarily) the positive root, not the negative one. Both [math](-3)^2 = 9[/math] and [math]3^2 = 9[/math]. However [math]\sqrt{9} = 3[/math], not -3. To get the negative root you need to negate the positive one. E.g. [math]-\sqrt{9} = -3[/math] gives the second root.What does it mean that the square root function arbitrarily selects the positive root? Well you could just as easily define a function, let’s call it N, that selects the negative root. Then we’d have[math] N(3^2) = N(9) = -3[/math][math]N(3)^2 = 3[/math]And[math]N((-3)^2) = N(9) = -3[/math][math]N(-3)^2 = (N(3)i)^2 = -3 [/math]So when we use a function that selects the negative root it works out fine in the negative case but not in the positive case. In this sense the choice of N isn’t any better or worse than the choice of [math]\sqrt{}[/math]. However [math]\sqrt{}[/math] does have some notational conveniences when it comes to positive numbers.[math]\sqrt{\sqrt{81}} = \sqrt{9} = 3[/math][math]N(N(81)) = N(-9) = -3 i[/math][math]-N(-N(81)) = -N(9) = -(-3) = 3 [/math]Simply chaining N’s will result in complex values due to the negative square root. It requires the additional specification of the positive root (-N(x)) to avoid this.Rather than specify a particular root it’s possible to specify both roots simultaneously using the [math]\pm[/math] symbol. Suppose [math]x^2 = 9[/math]. Then [math]x = \pm 3[/math] gives both solutions simultaneously. Note that some care needs to be taken with this notation. Specifically one may need to take care when negating the equations (sometimes written as [math]\mp[/math]) or incorporating multiple [math]\pm[/math] signs.
Absolute Value Inequality word problem. PLEASE HELP!!?
2.) Write an absolute value inequality that contains <= and has no solution. 3.)Devon tosses a horseshoe at a stake 30 feet away. The horseshoe lands no more than 3 feet away from the stake. (a) Write an absolute value inequality that represents the range of distances that the horseshoe travels.(b) Solve the inequality.
Use absolute value notation to describe the situation?
I'm not entirely sure what the question is so I'm going to fill in the blanks in a way that makes sense to me 1.|x-5|<=3 2.|y|>=6 I am assuming that these conditions specify the domain and range respectively so for our bottom functions y(assumed) =x^2-3x+4 (a) |-2-5| = 7 > 3 so -2 is not in your domain (b) |2-5| = 3 so 2 is in you domain now 2^2-3(2)+4 = 2 < 6 so it is not in your range. If I understand the question correctly it is not possible to evaluate either (a) or (b) with the above restrictions.