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2 years ago

Please please help me!!!

Mathematics

1 answer:

Nuetrik [128]2 years ago

3 0

Answer:

Step-by-step explanation:

The square means a $90$ degree angle.

We know that all the angles in a triangle add to $180$ degrees. So, we can build an equation for x: $(2x+1)+(5x+5)+(90)=180$

Combining like terms on the left side gives $7x+96=180$

Subtracting $96$ from both sides gives $7x=84$

Dividing both sides by $7$ gives $\boxed{x=12}$

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What is an equation of the line that passes through the points (2,5) and

harkovskaia [24]

Y-5=-1/2(X-2)
the simplified version of this equation (so you can plug into a calculator) is
Y=-1/2x+6

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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.

6 0

3 years ago

Guys, please help me with this question.

Elodia [21]

Answer:

Step-by-step explanation:

We are asked to perform the subtraction

x² + xy - 3y² - (5x² - xy + y²) ← distribute the parenthesis by - 1

= x² + xy - 3y² - 5x² + xy - y² ← collect like terms

= (x² - 5x²) + (xy + xy) + (- 3y² - y² )

= - 4x² + 2xy - 4y²

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3 years ago

Find de product of 15 and 107

Andru [333]

Answer:

$15 \times 107 = 1605$

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3 years ago

Read 2 more answers

Write a rule for the linear function in the table.

JulijaS [17]

Answer:

I guess that you want to know the transformations:

We start with:

f(x) = y = 4*x + 3

a)the transformed function is:

f(x) = y = -4*x - 3

So the sign changed.

This means that we go from (x, y) to (x, - y)

This is a reflection over the x-axis which changes the sin of the y component.

b) Now we go to f(x) = 4*x + 3

So the coefficient in the leading term changed.

This is a horizontal contraction:

A horizontal contraction of factor K for the function g(x) is: g(K*x)

In our case, we have:

f(K*x) = 4*(k*x) + 3 = x + 3

4*k*x = x

4*k = 1

k = 1/4

Then the transformation is an horizontal contraction of scale factor 1/4.

6 0

3 years ago