Ask question

Ask question

All categories

3 years ago

6

What is the area of a rectangle that is \dfrac1{2} 2 1 start fraction, 1, divided by, 2, end fraction of a meter long and \dfr

ac38 8 3 start fraction, 3, divided by, 8, end fraction of a meter wide?

1 answer:

sashaice [31]3 years ago

5 0

Answer:

$Area = \frac{3}{16} m^2$

Step-by-step explanation:

Given

Shape: Rectangle

$Length = \frac{1}{2}\ m$

$Width = \frac{3}{8}\ m$

Required

Determine the area of the rectangle;

Area is calculated as thus;

$Area = Length * Width$

Substitute values for Length and Width

$Area = \frac{1}{2} * \frac{3}{8}$

$Area = \frac{1 * 3}{2 * 8}$

$Area = \frac{3}{16}$

<em>Hence, the area of the rectangle is </em> $\frac{3}{16}m^2$ <em />

You might be interested in

Help me plz with this question!

sashaice [31]

I hope this help in some sort of way you can message me if you need more help

5 0

3 years ago

Identify the polygon with vertices A(5,0), B(2,4), C(−2,1), and D(1,−3), and then find the perimeter and area of the polygon. HE

inn [45]

Answer:

Part 1) The polygon is a square

Part 2) The perimeter is equal to $20\ units$

Part 3) The area is equal to $25\ units^{2}$

Step-by-step explanation:

we have

$A(5,0), B(2,4), C(-2,1),D(1,-3)$

Plot the points

see the attached figure

we know that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Find the distance AB

$A(5,0),B(2,4)$

substitute in the formula

$d=\sqrt{(4-0)^{2}+(2-5)^{2}}$

$d=\sqrt{(4)^{2}+(-3)^{2}}$

$d=\sqrt{25}$

$AB=5\ units$

Find the distance BC

$B(2,4), C(-2,1)$

substitute in the formula

$d=\sqrt{(1-4)^{2}+(-2-2)^{2}}$

$d=\sqrt{(-3)^{2}+(-4)^{2}}$

$d=\sqrt{25}$

$BC=5\ units$

Find the distance CD

$C(-2,1),D(1,-3)$

substitute in the formula

$d=\sqrt{(-3-1)^{2}+(1+2)^{2}}$

$d=\sqrt{(-4)^{2}+(3)^{2}}$

$d=\sqrt{25}$

$CD=5\ units$

Find the distance AD

$A(5,0),D(1,-3)$

substitute in the formula

$d=\sqrt{(-3-0)^{2}+(1-5)^{2}}$

$d=\sqrt{(-3)^{2}+(-4)^{2}}$

$d=\sqrt{25}$

$AD=5\ units$

we have that

AB=BC=CD=AD

Find the distance BD (diagonal)

$B(2,4),D(1,-3)$

substitute in the formula

$d=\sqrt{(-3-4)^{2}+(1-2)^{2}}$

$d=\sqrt{(-7)^{2}+(-1)^{2}}$

$BD=\sqrt{50}\ units$

<em>Verify if the polygon is a square</em>

If the triangle BDA is a right triangle, then the polygon is a square

Applying the Pythagoras theorem

$BD^{2}=AD^{2}+AB^{2}$

substitute

$(\sqrt{50})^{2}=5^{2}+5^{2}$

$50=50$ -----> is true

so

The triangle BDA is a right triangle

therefore

The polygon is a square

<em>Find the Area of the polygon</em>

The area of a square is equal to

$A=b^{2}$

we have

$b=5\ units$

$A=5^{2}=25\ units^{2}$

<em>Find the perimeter of the polygon</em>

The perimeter of a square is equal to

$P=4b$

we have

$b=5\ units$

$P=4(5)=20\ units$

6 0

3 years ago

The perimeter of a rectangle is 36m. If its length is increased by 1m and the width is increased by 2m, its area will increase b

maw [93]

Answer:

80 m^2

Step-by-step explanation:

The given information lets you write two equations involving length (x) and width (y).

2(x +y) = 36 . . . . the perimeter is 36 m
(x+1)(y+2) -xy = 30 . . . . increasing the length and width increases area

The second of these equations simplifies to another linear equation, giving a system of linear equations easily solved.

xy +y +2x + 2 -xy = 30

2x +y = 28 . . . . . . . subtract 2

Dividing the first equation by 2 gives

x +y = 18

and subtracting this from the above equation gives ...

(2x +y) -(x +y) = 28 -18

x = 10

Then

y = 18 -10 = 8

The area of the original rectangle is xy = 10·8 = 80 m^2.

4 0

3 years ago

2x^2+5x+3/x^2-3x-4 divided by 4x^2+2x-6/x^2-8x+16

Masja [62]

Answer: $\frac{x-4}{2x-2}$

Step-by-step explanation:

$\frac{2x^2+5x+3}{x^2-3x-4}$ divided by $\frac{4x^2+2x-6}{x^2-8x+16}$ is the same thing as multiplying $\frac{2x^2+5x+3}{x^2-3x-4}$ by $\frac{x^2-8x+16}{4x^2+2x-6}$ .

The equation we get is:

$\frac{2x^2+5x+3}{x^2-3x-4}*\frac{x^2-8x+16}{4x^2+2x-6}$

With this equation, we can factor each equation:

$\frac{(x+1)(2x+3)}{(x+1)(x-4)} *\frac{(x-4)(x-4)}{2(x-1)(2x+3)}$

We can cancel out like terms since they would be dividing each other:

$\frac{x-4}{2x-2}$

8 0

3 years ago

(8.5-2x)(11-2x)(x) what is the approximate value of x that would allow you to construct an

Wittaler [7]

The largest volume possible from one piece of paper for open-top box is 64.296 cubic unit.

<h3>What is meant by the term maxima?</h3>

The maxima point on the curve will be the highest point within the given range, and the minima point will be the lowest point just on curve.
Extrema is the product of maxima and minima.

For the given question dimensions of open-top box;

The volume is given by the equation;

V = (8.5-2x)(11-2x)(x)

Simplifying the equation;

V = x(4x² - 39x + 93.5)

Differentiate the equation with respect to x using the product rule.

dV/dx = x(8x -39) + (4x² - 39x + 93.5)

dV/dx = 8x² - 39x + 4x² - 39x + 93.5

dV/dx = 12x² - 72x + 93.5

Put the Derivative equals zero to get the critical point.

12x² - 72x + 93.5 = 0.

Solve using quadratic formula to get the values.

x = 4.1 and x = 1.9

Put each value of x in the volume to get the maximum volume;

V(4.1) = 4.1(4(4.1)² - 39(4.1) + 93.5)

V(4.1) = 3.44 cubic unit.

V(1.9) = 1.9(4(1.9)² - 39(1.9) + 93.5)

V(1.9) = 64.296 cubic unit. (largest volume)

Thus, the largest/maximum volume possible from one piece of paper for open-top box is 64.296 cubic unit.

To know more about the maxima, here

brainly.com/question/17184631

#SPJ1

4 0

1 year ago