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nadya68 [22]
3 years ago
14

At 4:00am the outside temperature was -28 by 4:00pm.it had risen 38 degrees what was the temperature at 4:00pm

Mathematics
2 answers:
barxatty [35]3 years ago
5 0

Answer:

10

Step-by-step explanation:

Georgia [21]3 years ago
4 0

Answer:

10° is the answer to the question

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What is value of 5.872 + (-0.84) ÷2.1
Lerok [7]

Answer: The answer is 5.472


Step-by-step explanation:

-0.84 ÷ 2.1 = -0.4

-0.4 + 5.872 =  <em>5.472</em>


6 0
3 years ago
How many real solutions exist for this system of equations?
Rom4ik [11]

Answer:

One

Step-by-step explanation:

Set each equations equal to each other

{x}^{2}  + 4 = 4x

{x}^{2}  - 4x + 4

Find the discrimant.

{ - 4 {}^{2} - 4(1)(4) } = 0

This means there is one real solution. Since the discramnt equal 0.

7 0
3 years ago
Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.
dalvyx [7]

Matching each equation of the plecewise function represented in the graph will be:

  • 2 < x < 3 = 3 - x
  • 0 < x < 2 = 1
  • x = 2 = x
  • 3 < x < 5 = 5 - x

<h3>How to illustrate the information?</h3>

It should be noted that a domain simply means the set of inputs that are accepted by the function.

In this case, the equation of the piecewise function represented is given.

The graph is attached.

Learn more about equations on:

brainly.com/question/2972832

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7 0
2 years ago
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
Hellpppp me please this test is my whole life and if I don’t pass I’m ruined for life so please HELPPPP MEEEEE
Elanso [62]

Answer:

expanded form: 5 * 5 * 5

value: 125

Step-by-step explanation:

3 0
3 years ago
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