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horrorfan [7]
2 years ago
11

What is the expression for 5.802

Mathematics
1 answer:
Reptile [31]2 years ago
8 0

Answer:

Step-by-step explanation:

Unsure of what you want.  If you want to express 5.802 in words, you get "five plus 802 thousandths."

If you want to break down 5.802 into an equivalent numerical expression, you get 5 + 8/10 + 2/1000, or

5 + 4/5 + 1/500

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How do i solve for <br> (2x/5)+1=13/5
NNADVOKAT [17]

Answer:

x = 4

Step-by-step explanation:

  1. \frac{2x}{5} +1 = \frac{13}{5}  
  2. \frac{2x+5}{5} = \frac{13}{5} divide both sides by 5
  3. 2x + 5 = 13 subtract 5 from both sides
  4. 2x = 8 divide both sides by 2
  5. x = 4
5 0
3 years ago
Which expression is equivalent to
Kruka [31]

Answer:

B) 1/x^2

Step-by-step explanation:

Simplify the following:

x^9/x^11

Hint: | For all exponents, a^n/a^m = a^(n - m). Apply this to x^9/x^11.

Combine powers. x^9/x^11 = x^(9 - 11):

x^(9 - 11)

Hint: | Evaluate 9 - 11.

9 - 11 = -2:

Answer:  x^(-2) = 1/(x^2)

5 0
2 years ago
PLEASE HELP ME!!! NO LINKS OR SITES, JUST THE ANSWER, THANK YOU!!
Helen [10]

Answer:

3.75

Step-by-step explanation:

12t = 45

t = 3.75

I hope this is correct

8 0
3 years ago
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statuscvo [17]

Number 1 is Rio Grande and number 2 is St. Lawrence River

Step-by-step explanation:

6 0
3 years ago
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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
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