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3 years ago

9

he equation of a line is y=2/3 x + 5. What is an equation of the line that is perpendicular to the given line and that passes th

rough the point (4,2)?

1 answer:

Levart [38]3 years ago

5 0

2/3 × m2 = - 1
m2 = - 3/2

2 = - 3/2(4) + c
4 = - 12 + 2c
2c = 16
c = 8
y = - 3/2x + 8

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Pete has $2.67 in dimes and pennies.If he has $1.16 in pennies how many dimes does Pete have

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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

Jobisdone [24]

<h3>Answer: 9.4 feet</h3>

Work Shown:

sin(angle) = opposite/hypotenuse

sin(22) = x/25

x = 25*sin(22)

x = 9.3651648353978

x = 9.4

Your calculator needs to be in degree mode. One way to check is to compute sin(30) and you should get 0.5 or 1/2.

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2 years ago

Light bulbs manufactured at a certain factory have a 3% probability of being defective. What is the probability that 5 out of a

Eddi Din [679]

There is a 3% chance of each light bulb being defective, and we only want 5 of them to be defective. The chance is then:

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3 years ago

Can someone help meeeeee

11Alexandr11 [23.1K]

Answer:

Area of shaded rectangle = 42 cm²

55%

Step-by-step explanation:

The length of the shaded area is equal to the length of the white rectangle.

Therefore, length of shaded rectangle = 7 cm

The width of the shaded rectangle is equal to the length of the white rectangle <u>minus</u> two widths of the white rectangle.

Therefore, width = 7 - (2 x 0.5) = 6 cm

Area of shaded rectangle = width x length

= 6 x 7

= 42 cm²

----------------------------------------------------------------------------------------------

Perimeter of a square = 4 × side length

If the perimeter of the square is 40 cm,
then the side length = 40 ÷ 4 = 10 cm

Area of a square = side length x side length

⇒ area of this square = 10 x 10 = 100 cm²

To determine the shaded area, calculate the areas of the 2 white triangles and subtract these from the area of the square.

Area of a triangle = 1/2 x base x height

⇒ area of left triangle = 1/2 x (10 - 7) x 10 = 15 cm²

⇒ area of right triangle = 1/2 x 10 x (10 - 4) = 30 cm²

Therefore, shaded area = 100 - 15 - 30 = 55 cm²

To calculate the percentage, divide the shaded area by the area of the square and multiply by 100:

(55 ÷ 100) × 100% = 55%

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2 years ago