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2 years ago

Math problem I need help with

Mathematics

2 answers:

Kruka [31]2 years ago

7 0

Answer:

31° is the answer thank you

elena55 [62]2 years ago

3 0

31 degrees Hopefully that helps!

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5 Vince draws the figures below. He says each figure has an area of 8 square units. Is he correct? Decide if each figure has an

igor_vitrenko [27]

Answer:ok so it is 1 A 2C 3 F 4H

Step-by-step explanation:

6 0

2 years ago

What is the slope of the line perpendicular to the given line?<br> 5x - 2y = 14

motikmotik

Answer:

$-\frac{2}{5}.$

Step-by-step explanation:

1) if to re-write the given equation into slope-interception form, then:

$y=\frac{5}{2}x-7;$

the slope of the given line is 5/2;

2) according to the rule of perpendicular, the required slope should be:

s*(5/2)=-1;

3) if to solve this equation (s*5/2=-1), then s=-2/5 or -0.4.

3 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

Your turn: Use the triangle given. What is the value of sin Θ?

LekaFEV [45]

You can eliminate D right away. The sine is NEVER greater than one. That's because the denominator of the fraction is always the hypotenuse.

The definition of the sine is
Sin(x) = Opposite / hypotenuse.
Sin(x) = 10/26 = 5/13 The fraction reduces.

The answer is B
Done.

3 0

2 years ago

What’s the answer to 0.462 / 0.2

Shalnov [3]

Answer: 2.31

Step-by-step explanation: 0.462 divided by 0.2 is 2.31. I knew it was going to be over 2 because 0.2 times 2 is 0.4.

5 0

2 years ago

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