All categories

3 years ago

Solve for x: 4|2x − 2| 10 = 34. (5 points) x = −4, x = 2 x = 4, x = −4 x = 2, x = −2 x = 4, x = −2

1 answer:

7 0

4|2x − 2| + 10 = 34
4|2x − 2| + 10 - 10 = 34 - 10
4|2x − 2| = 24
|2x − 2| = 24/4 = 6
2x - 2 = 6 or 2x - 2 = -6
2x = 6 + 2 or 2x = -6 + 2
2x = 8 or 2x = -4
x = 8/2 or x = -4/2
x = 4 or x = -2

To check for extraneous solutions:
For x = 4: 
4|2(4) − 2| + 10 = 34
4|8 − 2| = 34 - 10 = 24
6 = 24/4 = 6 (solution is valid)

For x = -2: 
4|2(-2) − 2| + 10 = 34
4|-4 − 2| = 34 - 10 = 24
|-6| = 6 = 24/4 = 6 (solution is valid) 

Therefore, solutions are x = 4 , x = -2

You might be interested in

Does anyone know this :-)

3241004551 [841]

<h2>Answer: -37</h2>

Step-by-step explanation:

i hope i helped

4 0

3 years ago

Read 2 more answers

If y = 4x – 1, which of the following sets represents possible inputs and outputs of the function, represented as ordered pairs?

Helga [31]

Hello!

To do this you will pick an x value, multiply it by the slope and add the y- intercept. Let's try 0.

y=4(0)-1
y=-1

So here, an ordered pair could be (0,-1)

We can try this again for different x-values.

y=4(1)-1
y=3

This gives us the ordered pair (1,3)

y=4(100)-1
y=399

This would give us the ordered pair (100,399)

I hope this helps!

6 0

3 years ago

This is word for word the problem I am working on in Trig. Its not a Trig question though!

denis-greek [22]

There will be 7 turkeys Xx

This is because:

52 divided by 2 = 26.

26 - 19 = 7.

-Hope this helps! Xx

8 0

3 years ago

How to find the vertex calculus 2What is the vertex, focus and directrix of x^2 = 6y

son4ous [18]

Solution:

Given:

$x^2=6y$

Part A:

The vertex of an up-down facing parabola of the form;

$\begin{gathered} y=ax^2+bx+c \\ is \\ x_v=-\frac{b}{2a} \end{gathered}$

Rewriting the equation given;

$\begin{gathered} 6y=x^2 \\ y=\frac{1}{6}x^2 \\ \\ \text{Hence,} \\ a=\frac{1}{6} \\ b=0 \\ c=0 \\ \\ \text{Hence,} \\ x_v=-\frac{b}{2a} \\ x_v=-\frac{0}{2(\frac{1}{6})} \\ x_v=0 \\ \\ _{} \\ \text{Substituting the value of x into y,} \\ y=\frac{1}{6}x^2 \\ y_v=\frac{1}{6}(0^2) \\ y_v=0 \\ \\ \text{Hence, the vertex is;} \\ (x_v,y_v)=(h,k)=(0,0) \end{gathered}$

Therefore, the vertex is (0,0)

Part B:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

$\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}$

Since the parabola is symmetric around the y-axis, the focus is a distance p from the center (0,0)

Hence,

$\begin{gathered} Focus\text{ is;} \\ (0,0+p) \\ =(0,0+\frac{3}{2}) \\ =(0,\frac{3}{2}) \end{gathered}$

Therefore, the focus is;

$(0,\frac{3}{2})$

Part C:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

Since the parabola is symmetric around the y-axis, the directrix is a line parallel to the x-axis at a distance p from the center (0,0).

Hence,

$\begin{gathered} Directrix\text{ is;} \\ y=0-p \\ y=0-\frac{3}{2} \\ y=-\frac{3}{2} \end{gathered}$

Therefore, the directrix is;

$y=-\frac{3}{2}$

3 0

1 year ago

What is a solution to a system of equations

zmey [24]

Answer:

A system of linear equations contains two or more equations e.g. y=0.5x+2 and y=x-2. The solution of such a system is the ordered pair that is a solution to both equations. ... The solution to the system will be in the point where the two lines intersect.

4 0

3 years ago