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2 years ago

9

Three hours ahead of ten forty in the morning

2 answers:

VashaNatasha [74]2 years ago

6 0

It's one forty PM in the afternoon (just look at a clock)

Maurinko [17]2 years ago

5 0

10 + 3 = 13
13-12 (noon) = 1
3 hours ahead of 10:40 in the morning is 1:40 in the afternoon. (13:40 in military)

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Express your final answer in simplest form 3/5 + -7/8

frez [133]

Answer:

- 0.5

Step-by-step explanation:

you first change the fractions numbers in to decimal form

so; 2/5 = 0.4

-7/8 = - 0.875

hence; 0.4 + (-0.875)

= - 0.475

in to its simplest form = - 0.5

the answer is - 0.5

5 0

2 years ago

A 200-gal tank contains 100 gal of pure water. At time t = 0, a salt-water solution containing 0.5 lb/gal of salt enters the tan

Artyom0805 [142]

Answer:

1) $\frac{dy}{dt}=2.5-\frac{3y}{2t+100}$

2) $y(t)=(50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}$

3) 98.23lbs

4) The salt concentration will increase without bound.

Step-by-step explanation:

1) Let $y$ represent the amount of salt in the tank at time t, where t is given in minutes.

Recall that: $\frac{dy}{dt}=rate\:in-rate\:out$

The amount coming in is $0.5\frac{lb}{gal}\times 5\frac{gal}{min}=2.5\frac{lb}{min}$

The rate going out depends on the concentration of salt in the tank at time t.

If there is $y(t)$ pounds of salt and there are $100+2t$ gallons at time t, then the concentration is: $\frac{y(t)}{2t+100}$

The rate of liquid leaving is is 3gal\min, so rate out is $=\frac{3y(t)}{2t+100}$

The required differential equation becomes:

$\frac{dy}{dt}=2.5-\frac{3y}{2t+100}$

2) We rewrite to obtain:

$\frac{dy}{dt}+\frac{3}{2t+100}y=2.5$

We multiply through by the integrating factor: $e^{\int \frac{3}{2t+100}dt }=e^{\frac{3}{2} \int \frac{1}{t+50}dt }=(50+t)^{\frac{3}{2} }$

to get:

$(50+t)^{\frac{3}{2} }\frac{dy}{dt}+(50+t)^{\frac{3}{2} }\cdot \frac{3}{2t+100}y=2.5(50+t)^{\frac{3}{2} }$

This gives us:

$((50+t)^{\frac{3}{2} }y)'=2.5(50+t)^{\frac{3}{2} }$

We integrate both sides with respect to t to get:

$(50+t)^{\frac{3}{2} }y=(50+t)^{\frac{5}{2} }+ C$

Multiply through by: $(50+t)^{-\frac{3}{2}}$ to get:

$y=(50+t)^{\frac{5}{2} }(50+t)^{-\frac{3}{2} }+ C(50+t)^{-\frac{3}{2} }$

$y(t)=(50+t)+ \frac{C}{(50+t)^{\frac{3}{2} }}$

We apply the initial condition: $y(0)=0$

$0=(50+0)+ \frac{C}{(50+0)^{\frac{3}{2} }}$

$C=-12500\sqrt{2}$

The amount of salt in the tank at time t is:

$y(t)=(50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}$

3) The tank will be full after 50 mins.

We put t=50 to find how pounds of salt it will contain:

$y(50)=(50+50)- \frac{12500\sqrt{2} }{(50+50)^{\frac{3}{2} }}$

$y(50)=98.23$

There will be 98.23 pounds of salt.

4) The limiting concentration of salt is given by:

$\lim_{t \to \infty}y(t)={ \lim_{t \to \infty} ( (50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }})$

As $t\to \infty$ , $50+t\to \infty$ and $\frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}\to 0$

This implies that:

$\lim_{t \to \infty}y(t)=\infty- 0=\infty$

If the tank had infinity capacity, there will be absolutely high(infinite) concentration of salt.

The salt concentration will increase without bound.

6 0

2 years ago

Factor using the magic X method:<br><br> x^2 + 6x + 5<br><br> SHOW WORK!!!

Schach [20]

Find values that add to 5 and multiply to 6
so the answer would be (x+5)(x+1)

4 0

2 years ago

Carlos is driving on a straight section of highway from Ashford to Lincoln. Ashford is at mile marker 333 and Lincoln is at mile

Arte-miy333 [17]

Distance between the two cities:

453 - 333 = 120 miles.

Rest area is 2/3 of the way:

120 x 2/3 = 240/3 = 80 miles.

Divide the miles to the rest stop by his speed:

80 miles/ 60 miles per hour = 1 and 1/3 hours as a fraction. 1.3333 as a decimal( round as needed.

( 1 hour and 20 minutes)

3 0

3 years ago

What is the mean, median, interquartile range, and standard deviation of 20, 20, 28, 28, 30, 30, 30, 36, 36, 40, 40

Kazeer [188]

Median is 30 range is 20 and that's all I'm gonna do sorry I have my own math hope I helped

4 0

3 years ago