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Mrac [35]
2 years ago
6

WHUR Radio uses 1/5 of its of its 10 hours of daytime broadcast for commercial and 1/2 for music. How much is left for news?

Mathematics
1 answer:
Nataly [62]2 years ago
8 0

Answer: 3 hours

Step-by-step explanation:

From the question, we are informed that WHUR Radio uses 1/5 of its of its 10 hours of daytime broadcast for commercial and 1/2 for music.

Hours used for commercial:

= 1/5 × 10

= 2 hours

Hours used for music:

= 1/2 × 10

= 5 hours

Hours left for news will be:

= 10 hours - 2 hours - 5 hours

= 3 hours

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2 years ago
WILLGIVEBRAINLIEST,THANKS,5STARS!<br> PLEASE HELP!<br> PLEASE EXPLAIN IN DEPTH!
bearhunter [10]
<h3>Answer:</h3>

x=2

<h3>Solution:</h3>
  • In order to isolate x, we should first of all take the square root of both sides.
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6 0
1 year ago
A swimming pool whose volume is 10 comma 000 gal contains water that is 0.03​% chlorine. Starting at tequals​0, city water conta
Katarina [22]

Answer:

C(60) = 2.7*10⁻⁴

t = 1870.72 s

Step-by-step explanation:

Let x(t) be the amount of chlorine in the pool at time t. Then the concentration of chlorine is  

C(t) = 3*10⁻⁴*x(t).

The input rate is 6*(0.001/100) = 6*10⁻⁵.

The output rate is 6*C(t) = 6*(3*10⁻⁴*x(t)) = 18*10⁻⁴*x(t)

The initial condition is x(0) = C(0)*10⁴/3 = (0.03/100)*10⁴/3 = 1.

The problem is to find C(60) in percents and to find t such that 3*10⁻⁴*x(t) = 0.002/100.  

Remember, 1 h = 60 minutes. The initial value problem is  

dx/dt= 6*10⁻⁵ - 18*10⁻⁴x =  - 6* 10⁻⁴*(3x - 10⁻¹)               x(0) = 1.

The equation is separable. It can be rewritten as dx/(3x - 10⁻¹) = -6*10⁻⁴dt.

The integration of both sides gives us  

Ln |3x - 0.1| / 3 = -6*10⁻⁴*t + C    or    |3x - 0.1| = e∧(3C)*e∧(-18*10⁻⁴t).  

Therefore, 3x - 0.1 = C₁*e∧(-18*10⁻⁴t).

Plug in the initial condition t = 0, x = 1 to obtain C₁ = 2.9.

Thus the solution to the IVP is

x(t) = (1/3)(2.9*e∧(-18*10⁻⁴t)+0.1)

then  

C(t) = 3*10⁻⁴*(1/3)(2.9*e∧(-18*10⁻⁴t)+0.1) = 10⁻⁴*(2.9*e∧(-18*10⁻⁴t)+0.1)

If  t = 60

We have

C(60) = 10⁻⁴*(2.9*e∧(-18*10⁻⁴*60)+0.1) = 2.7*10⁻⁴

Now, we obtain t such that 3*10⁻⁴*x(t) = 2*10⁻⁵

3*10⁻⁴*(1/3)(2.9*e∧(-18*10⁻⁴t)+0.1) = 2*10⁻⁵

t = 1870.72 s

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