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3 years ago

| x-3 | < x-3 can someone give a step by step process on how to do this?

Mathematics

1 answer:

statuscvo [17]3 years ago

4 0

Answer:

There are no solutions to the inequality.

Step-by-step explanation:

|x - 3| < x – 3

1. Separate the inequality into two separate ones.

(1) x – 3 < x – 3

(2) x – 3 < -(x – 3)

2. Solve each equation separately

(a) Equation (1)

$\begin{array}{rcl}x - 3 & < & x - 3\\x & < & x\\\end{array}\\\text{This is impossible. No solutions exist.}$

(b) Equation (2)

$\begin{array}{rcl}x - 3 & < & -(x - 3)\\x - 3 & < & -x + 3\\x &$

For example, if x = 0, we get

|0 - 3| < 0 - 3 or

3 < -3

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◆ Quadratic Equations ◆<br>Please help !

Mila [183]

I'm sure there's an easier way of solving it than the way I did, but I'm not sure what it could be. Never dealt with a problem like this before.

Anyway, I just plugged in and tested. Chose random values for a, b, c, and d, which follow the rule 0 < a < b < c < d:

a = 1
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$\sf ax^2+(1-a(b+c))x+abc-d)$

$\sf 1x^2+(1-1(2+3))x+(1)(2)(3)-(4))$

Simplify into standard form:

$\sf x^2+(1-1(5))x+6-4$

$\sf x^2+(1-5)x+2$

$\sf x^2-4x+2$

Use the quadratic formula to solve:

$\sf x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

For functions in the form of $\sf ax^2+bx+c$ . So in this case:

a = 1
b = -4
c = 2

Plug them in:

$\sf x=\dfrac{4\pm\sqrt{(-4)^2-4(1)(2)}}{2(1)}$

Solve for 'x':

$\sf x=\dfrac{4\pm\sqrt{16-8}}{2}$

$\sf x=\dfrac{4\pm\sqrt{8}}{2}$

$\sf x\approx\dfrac{4\pm 2.83}{2}$

$\sf x\approx 0.59,3.41$

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4 years ago

Jenna found a pair of sneaker that cost $64.She has a 15% off coupon. How much will she pay for the sneakers?(without Tax?)

S_A_V [24]

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4 years ago

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katrin [286]

Answer:

Step-by-step explanation:

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3 years ago

Need help with homework

motikmotik

We have two points describing the diameter of a circumference, these are:

$\begin{gathered} A=(-12,-4) \\ B=(-4,-10) \end{gathered}$

Recall that the equation for the standard form of a circle is:

$(x-h)^2+(y-k)^2=r^2$

Where (h,k) is the coordinate of the center of the circle, to find this coordinate, we find the midpoint of the diameter, that is, the midpoint between points A and B.

For this we use the following equation:

$M=(\frac{x_1+x_2_{}_{}}{2},\frac{y_1+y_2}{2})$

Now, we replace and solve:

$\begin{gathered} M=(\frac{-12+(-4)}{2},\frac{-4+(-10)}{2} \\ M=(\frac{-12-4}{2},\frac{-4-10}{2}) \\ M=(\frac{-16}{2},\frac{-14}{2}) \\ M=(-8,-7) \end{gathered}$

The center of the circle is (-8,-7), so:

$\begin{gathered} h=-8 \\ k=-7 \end{gathered}$

On the other hand, we must find the radius of the circle, remember that the radius of a circle goes from the center of the circumference to a point on its arc, for this we use the following equation: