Quick anyone know the answer to this question?

What is the answer to the question?

Mumz [18]

Answer:

The answer is 4 zeros

Step-by-step explanation:

(6×5)×10³

30×1000=30000

Therefore 4 zeros in 30000

8 0

3 years ago

Read 2 more answers

Autumn found a wallet containing $ 900 in the street. She returned the wallet to its owner, who gave her a $ 180 reward. What pe

Murljashka [212]

Answer:

$\frac{180}{900} = 0.2$ × $100 =20$

Autumn got a 20% reward of the $900 she found.

4 0

2 years ago

A vase in the shape of an oblique cylinder has the dimensions shown. What is the volume of the vase?

fenix001 [56]

The image is missing so i have attached it.

Answer:

Volume = 1.5 litres

Step-by-step explanation:

Using pythagoras theorem, we can get the height (h) of the cylinder

14² + h² = 17²

h² = 289 - 196

h = √93

Now, volume of a cylinder is;

V = πr²h

In the image, r = diameter/2 = 14/2 = 7cm

Thus,

V = π × 7² × √93

V = 1485 cm³

Now, 1 litre = 1000 cm³

Thus, volume = 1485/1000 = 1.485 litres ≈ 1.5 litres

5 0

2 years ago

Read 2 more answers

Which inequality is equivalent to absolute value x-4 >9

hoa [83]

Answer:

x> 13

Step-by-step explanation:

x-4 >9

Add 4 to each side

x-4+4 >9+4

x> 13

7 0

3 years ago

A product can be made in sizes huge, average and tiny which yield a net unit profit of $14, $10, and$5, respectively. Three cent

navik [9.2K]

Answer:

The model is:

z = 14* X₁₁ + 14*X₁₂ + 5*X₁₃ + 10*X₂₁ + 10*X₂₂ + 10*X₂₃ + 5*X₃₁ + 5*X₃₂ + 5*X₃₃ to maximize

Subject to:

First center X₁₁ + X₂₁ + X₃₁ ≤ 550

Second center X₁₂ + X₂₂ + X₃₂ ≤ 750

Third center X₁₃ + X₂₃ + X₃₃ ≤ 275

22* X₁₁ + 16* X₂₁ + 9*X₃₁ ≤ 11000

22* X₁₂ + 16* X₂₂ + 9*X₃₂ ≤ 2700

22*X₁₃ + 16* X₂₃ + 9*X₃₃ ≤ 3400

X₁₁ + X₁₂ + X₁₃ ≤ 710

X₂₁ + X₂₂ + X₂₃ ≤ 900

X₃₁ + X₃₂ + X₃₃ ≤ 350

2700*(X₁₁ + X ₂₁ + X₃₁) - 11000*(X₁₂ + X₂₂ + X₃₂) = 0

3400*(X₁₁ + X ₂₁ + X₃₁) - 11000*( ( X₁₃ + X₂₃ + X₃₃) = 0

Xij >= 0

Step-by-step explanation:

Let´s call Xij product size i produced in center j

According to this, we get the following set of variable

X₁₁ product size huge produced in center 1

X₁₂ product size huge produced in center 2

X₁₃ product size huge produced in center 3

X₂₁ product size average produced in center 1

X₂₂ product size average produced in center 2

X₂₃ product size average produced in center 3

X₃₁ product size-tiny produced in center 1

X₃₂ product size-tiny produced in center 2

X₃₃ product size-tiny produced in center 3

Then Objective function is

z = 14* X₁₁ + 14*X₁₂ + 5*X₁₃ + 10*X₂₁ + 10*X₂₂ + 10*X₂₃ + 5*X₃₁ + 5*X₃₂ + 5*X₃₃

Constrains

Center capacity

1.- First center X₁₁ + X₂₁ + X₃₁ ≤ 550

2.- Second center X₁₂ + X₂₂ + X₃₂ ≤ 750

3.- Third center X₁₃ + X₂₃ + X₃₃ ≤ 275

Water available

1.- 22* X₁₁ + 16* X₂₁ + 9*X₃₁ ≤ 11000

2.- 22* X₁₂ + 16* X₂₂ + 9*X₃₂ ≤ 2700

3.- 22*X₁₃ + 16* X₂₃ + 9*X₃₃ ≤ 3400

Demand constrain

Product huge

X₁₁ + X₁₂ + X₁₃ ≤ 710

Product average

X₂₁ + X₂₂ + X₂₃ ≤ 900

Product tiny

X₃₁ + X₃₂ + X₃₃ ≤ 350

Fraction SP/CC must be the same

First and second centers fraction SP/CC

(X₁₁ + X ₂₁ + X₃₁)/ 11000 = (X₁₂ + X₂₂ + X₃₂)/ 2700

2700*(X₁₁ + X ₂₁ + X₃₁) - 11000*(X₁₂ + X₂₂ + X₃₂) = 0

First and third centers fraction SP/CC

(X₁₁ + X ₂₁ + X₃₁)/ 11000 = ( X₁₃ + X₂₃ + X₃₃)/ 3400

3400*(X₁₁ + X ₂₁ + X₃₁) - 11000*( ( X₁₃ + X₂₃ + X₃₃) = 0

The model is:

z = 14* X₁₁ + 14*X₁₂ + 5*X₁₃ + 10*X₂₁ + 10*X₂₂ + 10*X₂₃ + 5*X₃₁ + 5*X₃₂ + 5*X₃₃

Subject to:

First center X₁₁ + X₂₁ + X₃₁ ≤ 550

Second center X₁₂ + X₂₂ + X₃₂ ≤ 750

Third center X₁₃ + X₂₃ + X₃₃ ≤ 275

22* X₁₁ + 16* X₂₁ + 9*X₃₁ ≤ 11000

22* X₁₂ + 16* X₂₂ + 9*X₃₂ ≤ 2700

22*X₁₃ + 16* X₂₃ + 9*X₃₃ ≤ 3400

X₁₁ + X₁₂ + X₁₃ ≤ 710

X₂₁ + X₂₂ + X₂₃ ≤ 900

X₃₁ + X₃₂ + X₃₃ ≤ 350

2700*(X₁₁ + X ₂₁ + X₃₁) - 11000*(X₁₂ + X₂₂ + X₃₂) = 0

3400*(X₁₁ + X ₂₁ + X₃₁) - 11000*( ( X₁₃ + X₂₃ + X₃₃) = 0

Xij >= 0

6 0

2 years ago