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Taya2010 [7]
3 years ago
14

One number is 6 more than another. The sum of the numbers is 30. Find the numbers

Mathematics
1 answer:
likoan [24]3 years ago
5 0

Answer:

The answers are 18 and 12

Step-by-step explanation:

You just start from 6 and 0 and keep on going like 7 and 1, 8 and 2, 9 and 3 etc. When I reached 18 and 12, the sum was 30, and it was 6 numbers apart.

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Radda [10]

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7 0
3 years ago
Read 2 more answers
Please help me with this I need this done!!!
mote1985 [20]

Answer:

1=1

2=4

3= <NTR=115

   <RTM=45

  <STN=45

4= A-124

    B-45

    C-45

    D-90

    E-124

5=15

6=?

Step-by-step explanation:

FGHJK

8 0
3 years ago
What is the answer to 6+9y-18=-3
lara [203]
6 + 9y - 18 = -3   Combine like terms (6 and -18)
      9y - 12 = -3   Add 12 to both sides
             9y = 9    Divide both sides by 9
               y = 1
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
3 years ago
Can someone please explain with answer?
ira [324]

Using: A^2+b^2=c^2

A&b= (√10)^2=10

10+10=c^2

20=c^2

(take the square root of 20)

c=2√5

Your answer is D

7 0
3 years ago
Read 2 more answers
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