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Natali [406]
3 years ago
13

Solve the following equation for x: 2(4x - 1) = 8x - 2

Mathematics
1 answer:
gulaghasi [49]3 years ago
4 0
I got 0=0 hope that helps :)
You might be interested in
The length of a rectangle is twice the width. The area is 162 yd? Find the length and the width.
Tresset [83]

Answer:

length = 2x = 2(9) = 18 yds

Step-by-step explanation:

Let width = x

Let length = 2x

Area = 162 yd2

 

length × width = Area

 

2x(x) = 162

2x2 = 162

 

Divide by 2 on both sides of equation.

 

x2 = 81

 

Square-root both sides of equation to undo the exponent.

 

x = √(81)

x = 9

 

Substitute this x value into the initial variables.

 

width = x = 9 yds

length = 2x = 2(9) = 18 yds

6 0
3 years ago
Consider the function f(x) = 3x − 2. Select True or False for each statement. The y-intercept is 2. A True B False The x-interce
iragen [17]

Answer: A false, B true, C true

Step-by-step explanation:

A) intercept with y axis

That means x=0

y= 3*0-2

y= -2 So A is false

B) intercept with x axis

That means y=0

0= 3x-2

2=3x

2/3= x B is true

C) The form of linear function is

f(x)= mx +n, where m is the slope

In this case m= 3 so C is true

8 0
3 years ago
Solve for t. <br> 4 (t + 1/4) = 3
tankabanditka [31]

Answer:

t=1/2

Step-by-step explanation:

3 0
3 years ago
What type of correlation is shown?<br> х<br> xx<br> Х<br> X
Sladkaya [172]
No type of correlation is shown but do you have a graph
8 0
2 years ago
Help thankss <br>answer is <br><img src="https://tex.z-dn.net/?f=%20%28%20-%2014.0%29" id="TexFormula1" title=" ( - 14.0)" alt="
lyudmila [28]
We have point P=(x_0,y_0)=(2,-4), so first calculate f'(x_0). There will be:

y=f(x)=2x^3-5x^2\\\\\\\\f'(x)=\big(2x^3-5x^2\big)'=\big(2x^3\big)'-\big(5x^2\big)'=2\big(x^3\big)'-5\big(x^2\big)'=\\\\=2\cdot3x^2-5\cdot2x=\boxed{6x^2-10x}\\\\\\\\f'(x_0)=f'(2)=6\cdot2^2-10\cdot2=6\cdot4-20=24-20=\boxed{4}

Now, we can write the equation of the normal line as:

y-y_0=-\dfrac{1}{f'(x_0)}(x-x_0)\\\\\\ y-(-4)=-\dfrac{1}{4}(x-2)\\\\\\y+4=-\dfrac{1}{4}x+\dfrac{1}{2}\\\\\\y=-\dfrac{1}{4}x+\dfrac{1}{2}-4\\\\\\\boxed{y=-\dfrac{1}{4}x-\dfrac{7}{2}}

Normal line (and every line) crosses x-axis when y = 0, so coordinates of A:

\boxed{y=0}\qquad\qquad\text{and}\\\\\\y=-\dfrac{1}{4}x-\dfrac{7}{2}\\\\\\&#10;0=-\dfrac{1}{4}x-\dfrac{7}{2}\\\\\\\dfrac{1}{4}x=-\dfrac{7}{2}\quad|\cdot4\\\\\\x=-\dfrac{7\cdot4}{2}\\\\\\x=-7\cdot2\\\\\\\boxed{x=-14}\\\\\\\boxed{A=(-14,0)}
8 0
3 years ago
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