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3 years ago

22i + 13 - (7i + 3+ 12i) + 16i - 25

Mathematics

2 answers:

loris [4]3 years ago

4 0

22i + 13 - 7i - 3 - 12i + 16i - 25
22i - 7i - 12i + 16i + 13 - 3 - 25
19i - 15 (Answer)

Sergio [31]3 years ago

3 0

Answer:

-15 + 19i

Step-by-step explanation:

For this equation, it is just a matter of simplifying.

So, first, distribute the - (negative) to the numbers in the parenthesis.

22i + 13 -7i - 3 - 12i + 16i - 25

Now simplify all like terms and write in standard form (a + bi).

-15 + 19i

This is your answer.

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Answer:

2.08333333333

Step-by-step explanation:

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3 years ago

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Imeldas jigsaw puzzle has 500 pieces. She estimates that she put togethe approximately 95 of the pieces.

tekilochka [14]

Answer:

19% done

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95/500=19

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3 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.

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3 years ago

Rena wrote all the integers from 1 to 19. How many digits did she write?

harkovskaia [24]

She wrote 19 digits

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3 years ago

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Help me solve this pls

ddd [48]

Answer:

$\frac{x - 10}{8}$

Step-by-step explanation:

$\frac{3x}{8} - \frac{x + 5}{4}$

The common denominator of both fractions is 8. Therefore,, divide the denominator of each fraction, then multiply what you get by the numerator of each fraction.

Thus:

$\frac{1(3x) - 2(x + 5)}{8}$

$\frac{3x - 2x - 10}{8}$

$\frac{x - 10}{8}$

5 0

3 years ago