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3 years ago

11

a rectangle poster board has an area of 3/8 square yard and a length of 1/2 yard. Given area equals length times width, how wide

is the poster board?

2 answers:

Lelu [443]3 years ago

8 0

Area = length × width

Length = 1/2 yd

Area = 3/8 yd

All you have to do is divide area by length to get width

3/8 ÷ 1/2

Keep 3/8
Change ÷ to ×
Flip 1/2 to 2/1

That equals 3/8 × 2/1 = 6/8 = 3/4

Width = 3/4 yd

~Aamira~

Hope this helped☺☺

mel-nik [20]3 years ago

5 0

Answer:

3/4 ya

Step-by-step explanation:

Keep 3/8

Change to times

Flip to 2/1

3/8 times 2/1=6/8=3/4

And there’s your anwser

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4 years ago

What is the equation for the plane illustrated below?

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Answer:

Hence, none of the options presented are valid. The plane is represented by $3 \cdot x + 3\cdot y + 2\cdot z = 6$ .

Step-by-step explanation:

The general equation in rectangular form for a 3-dimension plane is represented by:

$a\cdot x + b\cdot y + c\cdot z = d$

Where:

$x$ , $y$ , $z$ - Orthogonal inputs.

$a$ , $b$ , $c$ , $d$ - Plane constants.

The plane presented in the figure contains the following three points: (2, 0, 0), (0, 2, 0), (0, 0, 3)

For the determination of the resultant equation, three equations of line in three distinct planes orthogonal to each other. That is, expressions for the xy, yz and xz-planes with the resource of the general equation of the line:

xy-plane (2, 0, 0) and (0, 2, 0)

$y = m\cdot x + b$

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Where:

$m$ - Slope, dimensionless.

$x_{1}$ , $x_{2}$ - Initial and final values for the independent variable, dimensionless.

$y_{1}$ , $y_{2}$ - Initial and final values for the dependent variable, dimensionless.

$b$ - x-Intercept, dimensionless.

If $x_{1} = 2$ , $y_{1} = 0$ , $x_{2} = 0$ and $y_{2} = 2$ , then:

Slope

$m = \frac{2-0}{0-2}$

$m = -1$

x-Intercept

$b = y_{1} - m\cdot x_{1}$

$b = 0 -(-1)\cdot (2)$

$b = 2$

The equation of the line in the xy-plane is $y = -x+2$ or $x + y = 2$ , which is equivalent to $3\cdot x + 3\cdot y = 6$ .

yz-plane (0, 2, 0) and (0, 0, 3)

$z = m\cdot y + b$

$m = \frac{z_{2}-z_{1}}{y_{2}-y_{1}}$

Where:

$m$ - Slope, dimensionless.

$y_{1}$ , $y_{2}$ - Initial and final values for the independent variable, dimensionless.

$z_{1}$ , $z_{2}$ - Initial and final values for the dependent variable, dimensionless.

$b$ - y-Intercept, dimensionless.

If $y_{1} = 2$ , $z_{1} = 0$ , $y_{2} = 0$ and $z_{2} = 3$ , then:

Slope

$m = \frac{3-0}{0-2}$

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

y-Intercept

$b = z_{1} - m\cdot y_{1}$

$b = 0 -\left(-\frac{3}{2} \right)\cdot (2)$

$b = 3$

The equation of the line in the yz-plane is $z = -\frac{3}{2}\cdot y+3$ or $3\cdot y + 2\cdot z = 6$ .

xz-plane (2, 0, 0) and (0, 0, 3)

$z = m\cdot x + b$

$m = \frac{z_{2}-z_{1}}{x_{2}-x_{1}}$

Where:

$m$ - Slope, dimensionless.

$x_{1}$ , $x_{2}$ - Initial and final values for the independent variable, dimensionless.

$z_{1}$ , $z_{2}$ - Initial and final values for the dependent variable, dimensionless.

$b$ - z-Intercept, dimensionless.

If $x_{1} = 2$ , $z_{1} = 0$ , $x_{2} = 0$ and $z_{2} = 3$ , then:

Slope

$m = \frac{3-0}{0-2}$

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

x-Intercept

$b = z_{1} - m\cdot x_{1}$

$b = 0 -\left(-\frac{3}{2} \right)\cdot (2)$

$b = 3$

The equation of the line in the xz-plane is $z = -\frac{3}{2}\cdot x+3$ or $3\cdot x + 2\cdot z = 6$

After comparing each equation of the line to the definition of the equation of the plane, the following coefficients are obtained:

$a = 3$ , $b = 3$ , $c = 2$ , $d = 6$

Hence, none of the options presented are valid. The plane is represented by $3 \cdot x + 3\cdot y + 2\cdot z = 6$ .

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See attachment

EXPLANATION

The given inequality is

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Multiply both sides of the second inequality by -1 and reverse the inequality sign.

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The graphical solution to this inequality is shown in the attachment.

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