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Naddik [55]
2 years ago
10

I need help with this please

Mathematics
1 answer:
Westkost [7]2 years ago
8 0

Answer:

This alot but ill help you with 2 of them to learn em

Step-by-step explanation:

1) 2x^2 +5 =167

Subtract the 5 from both sides to get the equation 2x² = 162

Then divide the 2 from both sides to get x² = 81, Then square root from both sides to get x= 9

6) 2(2y +4)²=72

Divide 2 from both sides to get (2y+4)²=36

Then square root that to get 2y+4 = 6

subtract 4 from both sides to get 2y =2

Then divide both sides from 2. Y=2

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To rent a jet ski at Hayden’s cost is $25 plus $3 per hour. At Austin’s, it costs $5 plus $8 per hour. At how many hours will th
Katen [24]
<u>At Hayden's:</u> 

(25+3)= \ 28+3 \ (2nd \ hour)=31+3 \ (3rd \ hour)=34+3 . . . \swarrow\swarrow \\ \ \Longrightarrow \Longrightarrow \Longrightarrow \Longrightarrow \Longrightarrow \Longrightarrow \Longrightarrow \Longrightarrow \Longrightarrow \Longrightarrow \Longrightarrow  \boxed{34+3=37}= \ (4rth \ Hour)

<u>At Austins:</u> 

(5+8)= \ (first \ hour) = 13+8= \ 21 = (2nd \ hour) \ 21+8=29= \ \swarrow \\ \longrightarrow \ 29=(3rd \ hour)=\boxed{ 29+8= (37)= \ ((Forth \ hour))}

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3 0
3 years ago
How to solve 2x2+50=100
Ugo [173]

Answer:

It would be 54.

Step-by-step explanation:

It would be 2 times 2, which is four, and then you would add by 50.

2 x 2 = 4

4 + 50 = 54

So, it will not be 100, it will be 54.

4 0
3 years ago
Read 2 more answers
Better Products, Inc., manufactures three products on two machines. In a typical week, 40 hours are available on each machine. T
Kaylis [27]

Answer:

z (max)  =  1250 $

x₁  = 25    x₂  =  0   x₃  =  25

Step-by-step explanation:

                                Profit $    mach. 1      mach. 2

Product 1     ( x₁ )       30             0.5              1

Product 2    ( x₂ )       50             2                  1

Product 3    ( x₃ )       20             0.75             0.5

Machinne 1 require  2 operators

Machine   2 require  1  operator

Amaximum of  100 hours of labor available

Then Objective Function:

z  =  30*x₁  +  50*x₂  +  20*x₃      to maximize

Constraints:

1.-Machine 1 hours available  40

In machine 1    L-H  we will need

0.5*x₁  +  2*x₂  + 0.75*x₃  ≤  40

2.-Machine 2   hours available  40

1*x₁  +  1*x₂   + 0.5*x₃   ≤  40

3.-Labor-hours available   100

Machine 1     2*( 0.5*x₁ +  2*x₂  +  0.75*x₃ )

Machine  2       x₁   +   x₂   +  0.5*x₃  

Total labor-hours   :  

2*x₁  +  5*x₂  +  2*x₃  ≤  100

4.- Production requirement:

x₁  ≤  0.5 *( x₁ +  x₂  +  x₃ )     or   0.5*x₁  -  0.5*x₂  -  0.5*x₃  ≤ 0

5.-Production requirement:

x₃  ≥  0,2 * ( x₁  +  x₂   +  x₃ )  or    -0.2*x₁  - 0.2*x₂ + 0.8*x₃   ≥  0

General constraints:

x₁  ≥   0       x₂    ≥   0       x₃     ≥   0           all integers

The model is:

z  =  30*x₁  +  50*x₂  +  20*x₃      to maximize

Subject to:

0.5*x₁  +  2*x₂  + 0.75*x₃  ≤  40

1*x₁  +  1*x₂   + 0.5*x₃       ≤  40

2*x₁  +  5*x₂  +  2*x₃        ≤  100

0.5*x₁  -  0.5*x₂  -  0.5*x₃  ≤ 0

-0.2*x₁  - 0.2*x₂ + 0.8*x₃   ≥  0

x₁  ≥   0       x₂    ≥   0       x₃     ≥   0           all integers

After 6 iterations with the help of the on-line solver AtomZmaths we find

z (max)  =  1250 $

x₁  = 25    x₂  =  0   x₃  =  25

6 0
3 years ago
alicia cashed her paycheck and used half of her earnings to take three friends to dinner. She spent $42.15 on dinner and she bro
Nana76 [90]
X - $42.15 > $20.00
<u>  + $42.15 + $42.15</u>
              x > $62.15
5 0
3 years ago
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