Find the area of the shaded polygon.

77julia77 [94]

Answer:

Wow thats pretty hard the only things i could guess was the first one could be 18+ 18=36 and then there 6X6=36. but neither of those are greater than 50. And both of them are less than 75.

Step-by-step explanation:

Is this for a class??

7 0

3 years ago

Which of the following expressions is equivalent to (y − 4)4?

irina [24]

Answer:

It's equivalent to the 3rd answer choice: y4 − 16y3 + 96y2 − 256y + 256

hope this helps! <3

6 0

2 years ago

Find the perimeter of the figure to the nearest hundredth

Ghella [55]

Answer: 33.13 in

Step-by-step explanation:

First, let's start with the perimeter of the semicircles.

Since there are two, it is easier to do because the perimeter of the two halves will add to one full circle; so we can just calculate without having to halve anything.

The formula for the perimeter (aka circumference) of a circle is 2πr

Plug the radius into the equation to get 2π4, or 8π

8π ≈ 25.13

The perimeter of the rectangle is 6(4) = 24.

We now have to subtract the diameters of the circles, as they are not on the outside of the figure and won't be counted in the total perimeter. From the 24, we subtract the two diameters, or 16 in.

24 - 16 = 8

The total perimeter then comes out to be 25.13 + 8, or 33.13 in

3 0

4 years ago

Your swimming pool containing 60,000 gal of water has been contaminated by 6 kg of a nontoxic dye that leaves a swimmer's skin a

Paul [167]

[a] Dye is removed from the pool at a rate of

(250 gal/min) * (q/60,000 g/gal) = -q/240 g/gal

where q denotes the amount of dye in the pool at time t. Clean water is pumped back into the pool, so no dye is being re-added.

So the net rate of change of the amount of dye in the pool is given by the differential equation,

$\dfrac{\mathrm dq}{\mathrm dt}=-\dfrac{q(t)}{240}$

with the intial value, q(0) = 6000 g (or 6 kg).

[b] The ODE above is separable as

$\dfrac{\mathrm dq}q=-\dfrac{\mathrm dt}{240}$

Integrate both sides to get

$\ln|q|=-\dfrac t{240}+C$

$e^{\ln|q|}=e^{-t/240+C}$

$\implies q(t)=e^{-t/240+C}=e^{-t/240}e^C=Ce^{-t/240}$

Now plug in the initial condition:

$6000=Ce^0\implies C=6000$

so the particular solution to the IVP is

$q(t)=6000e^{-t/240}$

[c] The acceptable concentration of the dy is 0.03 g/gal, which in a pool containing 60,000 gal of water corresponds to

(0.03 g/gal) * (60,000 gal) = 1800 g = 1.8 kg

of dye. Find the time t when this occurs:

$1800=6000e^{-t/240}\implies0.3=e^{-t/240}$

$\implies\ln0.3=-\dfrac t{240}$

$\implies t=-240\ln0.3\approx288.953$

so the amount of dye in the pool is within the acceptable tolerance after about 289 min have passes, or about 4.82 hrs. So no, the filtration system is not up to the task.

8 0

3 years ago

Read 2 more answers

Mr. Gin purchased 3 pounds of dough and 1 pound of sauce.

Diano4ka-milaya [45]

Djshfjxfngusjtjvjsjrjvhsntjxnejgisjf

3 0

3 years ago