Lionardo melly true to his name,

The width of the rectangle is 0.6 meters less than the length. The perimeter of a rectangle is 60.8 meters. Find the dimensions

Anvisha [2.4K]

Answer:

14+=+2L+%2B+2%28L-3%29 Simplify and solve for L

14+=+2L+%2B+2L+-+6 Combine like-terms.

14+=+4L-6 Add 6 to both sides.

20+=+4L Divide both sides by 4.

5+=+L The length is 5 meters.

W+=+L-3

W+=+5-3

W+=+2meters.

The width is 2 meters.

P+=+2L%2B2W

P+=+2%285%29%2B2%282%29

P+=+10%2B4

P+=+14meters.

Step-by-step explanation:

4 0

3 years ago

What is 0.25x20 show and explain your answer

igomit [66]

Answer:

5

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Mrs. Smith is shopping for a toy chest to go in her kids' playroom. She looks at the options shown.

AfilCa [17]

The volume of a cuboid toy chest is equal to their product of the length,

width, and height.

The correct option for the toy chest she should purchase is; Only toy chest A will provide the necessary storage space.

Reasons:

The dimensions of the toy chest are;

Toy chest A;

Length = 5 feet

Width = 4 feet

Height = 2 feet

Toy chest B;

Length = 4 feet

Width = 2 feet

Height = 4 feet

Volume of Toy chest A = 5 ft. × 4 ft. × 2 ft. = 40 ft.³

Volume of Toy chest B = 4 ft. × 2 ft. × 4 ft. = 32 ft.³

The volume of the toy chest Mrs. Smith needs = 35 ft.³

The toy chest Mrs. Smith should purchase is Toy chest A, that has a volume of 40 ft.³, which can store items that with a volume of 35 ft.³

The correct option is; Only toy chest A will provide the necessary storage space

Possible question options obtained from a similar question found online are;

Either toy chest have the storage space needed

Neither has the storage space needed

Only toy chest A

Only toy chest B

Learn more about calculating volumes of various shapes here:

brainly.com/question/3789826

8 0

2 years ago

Given circle and circle with radii of 6cm and 4cm respectively.

noname [10]

Solution :

Given that :

The radius of circle A = 6 cm

The radius of circle C = 4 cm

In circle A

$\angle EAF = \theta = 140^\circ$

The length of arc EF = $2 \pi r \times \frac{\theta}{360^\circ}$

$=2 \times 3.14 \times 6 \times \frac{140^\circ}{360^\circ}$

= 14.653 cm

In circle C

$\angle GCH = \theta = 140^\circ$

The length of arc GH = $2 \pi r \times \frac{\theta}{360^\circ}$

$=2 \times 3.14 \times 4 \times \frac{140^\circ}{360^\circ}$

= 9.769 cm

Therefore,

The length of EF is 14.653 cm

The length of GH is 9.769 cm

The length of EF is 1.5 times the length of GH

i.e. 14.653 = 1.5 x 9.769

14.653 = 14.653

Hence proved.

6 0

3 years ago

The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2). On a coordinate plane, line A B has points (4, 1) and (n

GarryVolchara [31]

Answer:

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

Step-by-step explanation:

Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).

To find: coordinates of vertex of the right angle

Solution:

Let C be point $(x,y)$

Distance between points $(x_1,y_1),(x_2,y_2)$ is given by $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$AC=\sqrt{(x-4)^2+(y-1)^2}\\BC=\sqrt{(x+1)^2+(y+2)^2}\\AB=\sqrt{(4+1)^2+(1+2)^2}=\sqrt{25+9}=\sqrt{34}$

ΔABC is a right angled triangle, suing Pythagoras theorem (square of hypotenuse is equal to sum of squares of base and perpendicular)

$34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ]$

Put $(x,y)=(-1,1)$

$34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34$

which is true. So, $(-1,1)$ can be a vertex

Put $(x,y)=(4,-2)$

$34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34$

which is true. So, $(4,-2)$ can be a vertex

Put $(x,y)=(1,1)$

$34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22$

which is not true. So, $(1,1)$ cannot be a vertex

Put $(x,y)=(2,-2)$

$34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22$

which is not true. So, $(2,-2)$ cannot be a vertex

Put $(x,y)=(4,-1)$

$34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30$

which is not true. So, $(4,-1)$ cannot be a vertex

Put $(x,y)=(-1,4)$

$34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70$

which is not true. So, $(-1,4)$ cannot be a vertex

So, possible points for the vertex are $(-1,1),(4,-2)$

7 0

3 years ago

Read 2 more answers