Ask question

Ask question

All categories

3 years ago

6

If one real number is equal to a second real number, and the second real number is equal to a third real number, then the first

real number must equal the third:
If a = b and b = c, then a = c

1 answer:

svet-max [94.6K]3 years ago

8 0

Answer: In particular, let’s focus our attention on the behavior of each graph at and around . 2 and x= -1 for x < 2. There are open circles at both endpoints (2, 1) and (-2, 1). The third is h (x) = 1 / (x-2)^2, in which the function curves asymptotically towards y=0 and x=2 in quadrants one and two."

Step-by-step explanation: I think this is the problem ur on

You might be interested in

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

3 years ago

Which equations support the fact that rational numbers are closed under addition?

PtichkaEL [24]

i think the answer is the second choice √9+√9=2√9

<em>Hoped this helped!</em>

3 0

3 years ago

Anna35 [415]

Increasing. For every week that goes by, Olivia's puppy is gaining one pound. 6-5= 1 14-13= 1. Gaining a pound every week makes the puppies weight increase. Since increase is the term for something is bigger than number or value than before, Olivia's puppies' weight is increasing.

I hope this helps!
~kaikers

6 0

3 years ago

At a game booth, a student gets a box of candy as the prize for winning a game. The boxes come in four colors: white, red, green

Rainbow [258]

I think the answer is D) 8 over 32 multiplied by 7 over 31
Hopefully that helped! :)

4 0

3 years ago

If you are making a scale drawing of a girrafe that is 6m tall, and the height of the girrafe in your drawing is 3cm, what is th

k0ka [10]

Answer:

1cm to 200cm

Step-by-step explanation:

l of giraffe = 6m = 600cm

l of giraffe drawing = 3cm

scale = drawing : real = 3 : 600 = 1 : 200

Therefore the scale is 1cm to 200cm

8 0

2 years ago