The clerk at the checkout counter earned $350 in 25 hours.

Among all the rectangles whose perimeters are 100 feet, find the dimensions of the one with the maximum area. explain step by st

kompoz [17]

We let x be the length and y be the width of the rectangle. Then,

Perimeter = 2x + 2y
100 = 2x + 2y
50 = x + y
y = 50 - x

Area = xy
A = x(50 - x)
A = 50x - x^2

We then take the derivative; set it equal to zero:

A ' = 50 - 2x
0 = 50 - 2x
2x = 50
x = 25
y = 50 - x
y = 50 - 25
y = 25

Therefore, the dimensions are 25 and 25.

7 0

2 years ago

A rectangle has a length of (x + 4) centimeters and a width of 12x centimeters. If the perimeter is 26 centimeters, what is the

STALIN [3.7K]

Answer:

The width of rectangle is $W=\frac{108}{13}\ cm$ or $8\frac{4}{13}\ cm$

Step-by-step explanation:

we know that

The perimeter of rectangle is equal to

$P=2(L+W)$

where

L is the length of rectangle

W is the width of rectangle

we have

$P=26\ cm\\L=(x+4)\ cm\\W=12x\ cm$

substitute

$26=2(x+4+12x)$

solve for x

$26=2(13x+4)$

$26=26x+8$

$26x=26-8$

$26x=18$

$x=\frac{18}{26}$

simplify

$x=\frac{9}{13}$

Find the width of rectangle

$W=12x\ cm$

substitute the value of x

$W=12(\frac{9}{13})=\frac{108}{13}\ cm$

Convert to mixed number

$\frac{108}{13}\ cm=\frac{104}{13}+\frac{4}{13}=8\frac{4}{13}\ cm$

7 0

2 years ago

What percent of 40 is 28.4?

zalisa [80]

71% of 40 is 28.4

Divide 28.4 by 40:
$\frac{28.4}{40} = 0.71$

Change the decimal to a percentage by multiplying by 100:
$0.71 \times 100 = 71$

6 0

3 years ago

Read 2 more answers

Pls help and show workings

andreev551 [17]

Answer: x = 11

Step-by-step explanation:

same as the last question, set them as equal and solve. here's my work:

5x - 1 = 4x + 10

5x - 1 (+ 1) = 4x + 10 (+ 1)

5x = 4x + 11

5x (- 4x) = 4x (- 4x) + 11

x = 11

8 0

2 years ago

A stadium has a a central rectangular area 125m long by 80m wide.there are two semi circular ends.a running track goes all the w

Fudgin [204]

a)The total inside length of the curved ends of the track will be 25.13 meters

b)The total distance around the inside of the track will be 275.13 meters

c)The area inside the track is 10050 square meters.

The space occupied by any two-dimensional figure in a plane is called the area. The space occupied by the circle in a two-dimensional plane is called the area of the circle.

Given that:-

A stadium has a central rectangular area 125m long by 80m wide. there are two semi-circular ends. a running track goes all the way around.

a) what is the total inside length of the curved ends of the track?

The length of the curved track means it is a perimeter of the circle so the radius of the semicircle will be 4 meters.

So the length will be = π r + πr = 2 πr

= 2 x π x 4 = 25.13

b) what is the total distance around the inside of the track?

The total distance of the track will be the sum of the length of the two semicircles and the longer side of the rectangle.

Total length = 125 + 125 + 25.13

= 275.13 meters

c) what is the area inside of the track?

The area will be equal to the area of the two semicircular tracks and the rectangular track.

Total Area = Area of semicircle + Area of semicircle + Area of rectangle

= ( π/2 ) r² + (π/2) r² + ( L x W )

= π r² + ( L x W)

= π ( 4 )² + ( 125 x 8 )

= 10050.26 square meters.

Therefore the total inside length of the curved ends of the track will be 25.13 meters. The total distance around the inside of the track will be 275.13 meters. The area inside the track is 10050 square meters.

To know more about an area follow

brainly.com/question/25292087

#SPJ1

3 0

1 year ago