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djyliett [7]
3 years ago
15

1/3(m-1)=-5 Please help

Mathematics
1 answer:
dezoksy [38]3 years ago
6 0

Answer:

m=-14 I hope this helps u

Step-by-step explanation:


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Determine whether the solution of -2x = -15 is<br> Greater than or less than -15
brilliants [131]

Positive numbers are always greater than -15. Therefore, it's a greater than.

-2x = - 15

Dividing both sides by -2, we get

x = 7.5

Negative cancels on both sides, so we get a positive equation.

Positive numbers are always greater than -15. Therefore, it's a greater than.

What is greater than or less than?

  • Greater than and less than are the comparison symbols.
  • When the number is bigger or smaller than the other, then greater than and less than symbols are used.
  • If the number is greater than the other, the greater than (>) symbol is used.
  • If the number is lesser than the other, the less than the (<) symbol is used.

To learn more about greater than and less than, visit: brainly.com/question/15746367

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5 0
1 year ago
A new playground is in the shape of a square. on a scale drawing, the vertices of the playground’s coordinates are: (6,0.5)
Lisa [10]
What is the question exactly? What is it you are trying to find?
4 0
3 years ago
What is the solution set for 5x−5=15, given the replacement set {0, 1, 2, 3, 4}?
True [87]

Answer:

x=4

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
Suppose brine containing 0.2 kg of salt per liter runs into a tank initially filled with 500 L of water containing 5 kg of salt.
Oliga [24]

Answer:

(a) 0.288 kg/liter

(b) 0.061408 kg/liter

Step-by-step explanation:

(a) The mass of salt entering the tank per minute, x = 0.2 kg/L × 5 L/minute = 1 kg/minute

The mass of salt exiting the tank per minute = 5 × (5 + x)/500

The increase per minute, Δ/dt, in the mass of salt in the tank is given as follows;

Δ/dt = x - 5 × (5 + x)/500

The increase, in mass, Δ, after an increase in time, dt, is therefore;

Δ = (x - 5 × (5 + x)/500)·dt

Integrating with a graphing calculator, with limits 0, 10, gives;

Δ = (99·x - 5)/10

Substituting x = 1 gives

(99 × 1 - 5)/10 = 9.4 kg

The concentration of the salt and water in the tank after 10 minutes = (Initial mass of salt in the tank + Increase in the mass of the salt in the tank)/(Volume of the tank)

∴ The concentration of the salt and water in the tank after 10 minutes =  (5 + 9.4)/500 = (14.4)/500 = 0.288

The concentration of the salt and water in the tank after 10 minutes = 0.288 kg/liter

(b) With the added leak, we now have;

Δ/dt = x - 6 × (14.4 + x)/500

Δ = x - 6 × (14.4 + x)/500·dt

Integrating with a graphing calculator, with limits 0, 20, gives;

Δ = 19.76·x -3.456 = 16.304

Where x = 1

The increase in mass after an increase in = 16.304 kg

The total mass = 16.304 + 14.4 = 30.704 kg

The concentration of the salt in the tank then becomes;

Concentration = 30.704/500 = 0.061408 kg/liter.

6 0
3 years ago
Can someone assist me with this, I'm having problems with this
nirvana33 [79]
It’s true!! Hope this helped!!!!
8 0
2 years ago
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