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3 years ago

What is the value of x in the equation 2(x+3)=4(x-1)

1 answer:

5 0

Firstly, expand the equation.
$2(x + 3) = 4(x - 1) \\ 2x + 6 = 4x - 4$

Next, move the variable x to the left and all the constant to the right.
$2x + 6 = 4x - 4 \\ 2x - 4x = - 4 - 6 \\ - 2x = - 10$

Then, solve for x.
$- 2x = - 10 \\ x = \frac{ - 10}{ - 2} \\ x = 5$

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What is the prime factorization of 135?

Brums [2.3K]

135 is a composite number.

135 = 1 x 135, 3 x 45, 5 x 27, or 9 x 15.

Factors of 135: 1, 3, 5, 9, 15, 27, 45, 135.

Prime factorization: 135 = 3 x 3 x 3 x 5, which can also be written 135 = 3³ x 5

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3 years ago

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slega [8]

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Alexei wants to hang a mirror in his boat and put a frame around it. The mirror and frame must have an area of 19.25 square feet

zhuklara [117]

Let

x---------> the thickness of the frame in feet

At-------> Area of the mirror and the frame in square feet

we know that

$At=19.25ft^{2}$ ------> equation $1$

$At=(3+2x)*(5+2x)\\ At=3*5+3*2x+5*2x+2x*2x\\ At=15+6x+10x+4x^{2} \\ At=4x^{2} +16x+15$ --------> equation $2$

equate equation $1$ and equation $2$

$19.25=4x^{2} +16x+15\\ 4x^{2} +16x+15-19.25=0\\ 4x^{2} +16x-4.25=0$

therefore

the answer is the option

$C. 4x2 + 16x - 4.25 = 0$

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Lady bird [3.3K]

48 inches cubed is the answer

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A square on a coordinate plane has the points A(0, 0), B(0, 5), C(5, 5) and D(5, 0). If Square ABCD is rotated about the origin

Georgia [21]

Answer:

B. 5 units

Step-by-step explanation:

Transformation is the movement of a point from its initial location to a new location. Types of transformation are reflection, dilation, rotation or translation.

If a point X(x,y) is rotated about the origin 180 degrees clockwise the new point is at X'(-x, -y).

If the square with vertices at A(0, 0), B(0, 5), C(5, 5) and D(5, 0) is rotated about the origin 180 degrees clockwise the new points are at A'(0, 0), B'(0, -5), C'(-5, -5), D'(-5, 0)

A square has four equal sides. The distance between two points $X(x_1,y_1)\ and\ Y(x_2,y_2)$ is:

$|XY|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Hence:

$|A'B'|=\sqrt{(0-0)^2+(-5-0)^2} =5$

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3 years ago