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3 years ago

Question 2, please :)

Mathematics

1 answer:

Dmitry_Shevchenko [17]3 years ago

5 0

The average rate of change between week 4 and week 12 would be + $0.75 per week.

Hope this helps!

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7. What is the value of x for the triangle drawn below?

slega [8]

Answer:

where is the triangle?

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4 years ago

a solid metal ball with a radius of 10 inches is melted and made into smaller spherical metal balls with a radius of 2 inches ea

Leto [7]

Answer:

5 i think

Step-by-step explanation:

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3 years ago

In ΔKLM, m = 160 cm, ∠M=145° and ∠K=28°. Find the length of l, to the nearest 10th of a centimeter.

Montano1993 [528]

Answer:

34.0 m

Step-by-step explanation:

The total angle sum in a triangle should be 180° hence the missing angle will be given by

180°-145°-28°=7°

Using sine rule

$\frac {m}{sin M}=\frac {l}{sin L}=\frac {k}{sin K}$

By substituting 160 for m, 145° for M and 7° for L then

$\frac {160}{sin 145^{\circ}}=\frac {l}{sin 7^{\circ}}$

Making l the subject then

$l=\frac {160 sin 7^{\circ}}{sin 145^{\circ}}=33.9956345990643m\\l\approx 34.0m$

3 0

3 years ago

A 500 gallon tank initially contains 200 gallons of water with 5 lbs of salt dissolved in it. Water enters the tank at a rate of

Lapatulllka [165]

Until the concerns I raised in the comments are resolved, you can still set up the differential equation that gives the amount of salt within the tank over time. Call it $A(t)$ .

Then the ODE representing the change in the amount of salt over time is

$\dfrac{\mathrm dA}{\mathrm dt}=\text{rate in}-\text{rate out}$
$\dfrac{\mathrm dA}{\mathrm dt}=\dfrac{2\text{ gal}}{1\text{ hr}}\times\dfrac{\frac15(1+\cos t)\text{ lbs}}{1\text{ gal}}-\dfrac{2\text{ gal}}{1\text{ hr}}\times\dfrac{A(t)\text{ lbs}}{500+(2-2)t}$
$\dfrac{\mathrm dA}{\mathrm dt}=\dfrac25(1+\cos t)-\dfrac1{250}A(t)$

and this with the initial condition $A(0)=5$

You have

$\dfrac{\mathrm dA}{\mathrm dt}+\dfrac1{250}A(t)=\dfrac25(1+\cos t)$
$e^{t/250}\dfrac{\mathrm dA}{\mathrm dt}+\dfrac1{250}e^{t/250}A(t)=\dfrac25e^{t/250}(1+\cos t)$
$\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/250}A(t)\right]=\dfrac25e^{t/250}(1+\cos t)$

Integrating both sides gives

$e^{t/250}A(t)=100e^{t/250}\left(1+\dfrac1{62501}\cos t+\dfrac{250}{62501}\sin t\right)+C$
$A(t)=100\left(1+\dfrac1{62501}\cos t+\dfrac{250}{62501}\sin t\right)+Ce^{-t/250}$

Since $A(0)=5$ , you get

$5=100\left(1+\dfrac1{62501}\right)+C\implies C=-\dfrac{5937695}{62501}$

so the amount of salt at any given time in the tank is

$A(t)=100\left(1+\dfrac1{62501}\cos t+\dfrac{250}{62501}\sin t\right)-\dfrac{5937695}{62501}e^{-t/250}$

The tank will never overflow, since the same amount of solution flows into the tank as it does out of the tank, so with the given conditions it's not possible to answer the question.

However, you can make some observations about end behavior. As $t\to\infty$ , the exponential term vanishes and the amount of salt in the tank will oscillate between a maximum of about 100.4 lbs and a minimum of 99.6 lbs.

5 0

4 years ago

A ball bearing is shaped like a sphere and has a diameter of 2.5 centimetres. What is the volume contained inside the ball beari

kvv77 [185]

To solve:

The diameter is already given (2.5), and it can be divided by 2 to find the radius.

2.5/2=1.25

Plug 1.25 and Pi into the equation for a sphere to find the volume:

(4/3)(1.25^3)(3.14)=8.177

Answer: About 8.18 cm^3

8 0

3 years ago

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