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4 years ago

Have a great wonderful fantastic amazingggg dayyyy!!!

Mathematics

2 answers:

ddd [48]4 years ago

8 0

Answer:

Step-by-step explanation:

To start you have to calculate the perimeter of the white box. 1 times 2 is 2. 4 times two is 8. 8 plus 2 is 10. Then you have to find the perimeter of the purple box. 12 times 2 is 24. 15 times 2 is 30. 24 plus 30 is 54. After that you have to subtract to get the final answer. 54 minus 10 in 44. 44 is the final answer.

SCORPION-xisa [38]4 years ago

6 0

Answer: 176m^2

Step-by-step explanation:

The smaller rectangle is taking up space among the larger one. So in order to find the area of this shaded region, you need to subtract that smaller one from the larger one.

12 x 15 = 180

1 x 4 = 4

180 - 4 = 176m^2

You might be interested in

2x-1/4(12x+4)=3 2x-1/4 (12x+4)=3

eduard

2x-1/4(12x+4)=3
⇒ 2x -1/4*(12x) -1/4*4= 3 (distributive property)
⇒ 2x -3x -1= 3
⇒ -x= 3+1
⇒ -x= 4
⇒ x= -4

Final answer: x=-4~

3 0

3 years ago

A sphere has a volume of 1,436. 03m3, what is the approximate length of its radius?.

makkiz [27]

The sphere with the volume 1436.03 m³ has 7m long radius.

To find out the approximate length of its radius by using the formula to calculate volume of sphere is:

V= 4πr³/3

The value of π is 22/7 or 3.14.

So, V = 4 × 3.14 × r³ / 3

Volume of the sphere is given which is 1436.03 m³.

Thus, r³ = 1436.03×3 / 4×3.14

r = ³√ 4308.09/12.56 = ³√343 = 7 m.

Hence, the radius of the given sphere is 7 m.

The cube root of a number is the factor that we multiply by itself three times to get that number.

For example, the cube root of 27 is 3.

To learn more about the sphere click here brainly.com/question/19868745

#SPJ4

3 0

2 years ago

There are 17 portable mini suites (a.k.a. cages) in a row at the paws and claws holiday pet resort. they are neatly labeled with

DiKsa [7]

Given that there are 17 portable mini suites (a.k.a. cages) in a row at the Paws and Claws Holiday Pet Resort. They are neatly labeled with their guests' names. There are 8 poodles and 9 tabbies. How many ways can the "suites" be arranged if: a) there are no restrictions.

b) cats and dogs must alternate.

c) dogs must be next to each other.

d) dogs must be next to each other and cats must be next to each other.

Part A:

If there are no restrictions, then the number of ways the "suites" can be arranged is given by:

$17!=17\times16\times15\times14\times13\times12\times11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1=355,687,428,096,000 \ ways$

Part B:

If cats and dogs must alternate, then the number of ways the "suites" can be arranged is given by:

$9!\times8!=9\times8\times7\times6\times5\times4\times3\times2\times1\times8\times7\times6\times5\times4\times3\times2\times1=14,631,321,600 \ ways$

Part C:

If dogs must be next to each other, then the group of dogs are taken as 1 object and there are now 10 objects to arrange. The number of ways to arrange 10 objects is given by:

$10!=10\times9\times8\times7\times6\times5\times4\times3\times2\times1=3,628,800
 \ ways$

Also, the group of dogs can be arranged in:

$8!=8\times7\times6\times5\times4\times3\times2\times1=40,320
 \ ways$

Therefore, the total number of ways the "suites" can be arranged is given by:

$10!\times8!=3,628,800\times40,320=146,313,216,000 \ ways$

Part D:

If dogs must be next to each other and cats must be next to each other, then the group of dogs are taken as 1 object and the group of cats are taken as 1 object, then there are now 2 objects to arrange.

The number of ways to arrange two objects is given by:

$2!=2\times1=2$

The number of ways the 9 dogs are to be arranged is given by:

$9!=9\times8\times7\times6\times5\times4\times3\times2\times1=362,880$

The number of ways the 8 cats are to be arranged is given by:
$8!=8\times7\times6\times5\times4\times3\times2\times1=40,320$

Therefore, the total number of ways the "suites" can be arranged is given by:

$2!\times9!\times8!=2\times362,880\times40,320=29,262,643,200 \ ways$

4 0

3 years ago

A pig gained 14 pounds, and it now weighs 45 pounds. Write 4 equations to represent p, the weight of the pig before gaining the

shusha [124]

Let the original weight of the pig be p.
p+14=45
p=31.

3 0

4 years ago

Need help please, thank you

faust18 [17]

M<ABX + m<CBX = 180

therefore, (14x + 70) + (20x + 8) = 180

Subtract 78 from both sides. Divide by 34.
x = 3

Plug in 3 for x.

m<ABX = 14(3) + 70 = 112
m<CBX =20(3) + 8 = 68

5 0

4 years ago

Have a great wonderful fantastic amazingggg dayyyy!!!​

Have a great wonderful fantastic amazingggg dayyyy!!!