All categories

4 years ago

Need help ASAP!!!!!!!!

Mathematics

1 answer:

andriy [413]4 years ago

7 0

Answer:

Step-by-step explanation

V=1/3BH= 1/3 X area of base X height

your answer should be 270

You might be interested in

What is the area of this figure?

Oxana [17]

Answer:

4 ssquare units

Step-by-step explanation:

3 0

3 years ago

Buffalo Bill currently weighs 202 lb. He wants to weigh 180 lbs for his reunion which is 35 days from today. How many pounds per

Nostrana [21]

Answer:

He must lose 4.4 pounds per week.

Step-by-step explanation:

Buffalo Bill currently weighs 202 lb. He wants to weigh 180 lbs for his reunion.

This means that he needs to lose 202 - 180 = 22 lb.

35 days from today.

Each week has 7 days. So this is 35/7 = 5 weeks from now.

How many pounds per week must he lose?

22 pounds in 5 weeks, that is 22/5 = 4.4 pounds per week.

4 0

3 years ago

HELPPPP <br> this is due today <br> WITH THE EQUATION

timofeeve [1]

60% more than regular size. 230.4.

5 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

Please Help

masya89 [10]

1) The first graph is the graph of f(x) traslated 5 units to the left, then the equation of this graph is: y=(x+5)^2

2) The second graph is the graph of f(x) traslated 5 units upward, then the equation of this graph is: y=x^2+5

3) The third graph is the graph of f(x) traslated 5 units downward, then the equation of this graph is: y=x^2-5

4) The fourth graph is the graph of f(x) traslated 5 units to the right, then the equation of this graph is: y=(x-5)^2

4 0

3 years ago