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3 years ago

The length of a rectangle is two times its width. Find the area if its perimeter is 7 1/3

Mathematics

1 answer:

Nikolay [14]3 years ago

8 0

Let x be the width

width = x

Length is twice its width

length = 2x

Perimeter = length + width + length + width

Perimeter = x + 2x + x + 2x = 6x

Given that perimeter = 7 1/3

6x = 7 1/3

x = 7 1/3 ÷ 6 = 1 2/9

So width = 1 2/9

Length = 1 2/9 x 2 = 2 4/9

Area = length x width

Area = 1 2/9 x 2 4/9

Area = 2 80/81 square cm

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3 years ago

A remote control car races down the street at 26 miles per hour. Two hours later a second remote control car races straight down

Neporo4naja [7]

Answer:

4 hours

Step-by-step explanation:

Given

Car 1:

$Speed = 26mph$

$Time = h$

Car 2:

$Speed = 52mph$

$Time = h - 2$ --- i.e. 2 hours after car 1

Required

Determine the time car 1 will meet car 2

To do this, we calculate the distance of both cars using:

$Distance = Speed * Time$

For Car 1:

$Distance = 26 * h$

$Distance = 26h$

For Car 2:

$x^{2} Distance = 52 * (h - 2)$

Open bracket

$Distance = 52h - 104$

When car 2 meets up with car 1, they are at the same distance.

So:

$26h = 52h - 104$

Collect Like Terms

$26h - 52h = -104$

$-26h = -104$

Solve for h

$h = \frac{-104}{-26}$

$h = 4$

<em>Car 2 will catch car 1; 4 hours from the start</em>

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3 years ago

Lamar is writing a coordinate proof to show that a segment from the midpoint of the hypotenuse of a right triangle to the opposi

Zanzabum

1. N is a midpoint of the segment KL, then N has coordinates

$\left(\dfrac{x_K+x_L}{2},\dfrac{y_K+y_L}{2} \right) =\left(\dfrac{0+2a}{2},\dfrac{2b+0}{2} \right) =(a,b).$

2. To find the area of △KNM, the length of the base MK is 2b, and the length of the height is a. So an expression for the area of △KNM is

$A_{KMN}=\dfrac{1}{2}\cdot \text{base}\cdot \text{height}=\dfrac{1}{2}\cdot 2b\cdot a=ab.$

3. To find the area of △MNL, the length of the base ML is 2a and the length of the height is b. So an expression for the area of △MNL is

$A_{MNL}=\dfrac{1}{2}\cdot \text{base}\cdot \text{height}=\dfrac{1}{2}\cdot 2a\cdot b=ab.$

4. Comparing the expressions for the areas you have that the area $A_{KMN}$ is equal to the area $A_{MNL}$ . This means that the segment from the midpoint of the hypotenuse of a right triangle to the opposite vertex forms two triangles with equal areas.