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The net of a cylinder is shown. The area of one base of the cylinder is 9Pi in.2. The circumference of a base is 6Pi in., and th

hram777 [196]

Answer:

9Pi + 9Pi + 6Pi(10) in.^2

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Find the perimeter or cm

kicyunya [14]

Answer:

The perimeter is 22.

Step-by-step explanation:

The perimeter formula (it sometimes varies) is 2(b+h)

b being base (in this example, 6)

h being height (in this example, 5)

So now, with substitution, we are left with:

2(6+5)

Which simplifies down to:

2(11)

22

8 0

3 years ago

The cattle at the Clinton Farm are fed 1/4 of a bale of hay each day. The horses are fed 2 times as much hay as the cattle. How

xz_007 [3.2K]

Answer:

1/2 of a bale

Step-by-step explanation:

1/4 times 2 equals 1/2

7 0

2 years ago

(PLEASE ANSWER: TIMED)The intensity, or loudness, of a sound can be measured in decibels (dB), according to the equation I(dB)=1

Mrac [35]

For this case we have the following equation:
$l(dB) = 10log( \frac{l}{lo} )$
We must replace the following value in the equation:
$l = 10^8lo$
Substituting we have:
$l(dB) = 10log( \frac{10^8lo}{lo} )$
Simplifying the given expression we have:
$l(dB) = 10log(10^8)$
Then, using logarithm properties in base 10, we can rewrite the expression:
$l(dB) = 10(8)$
Finally, making the product, the result is:
$l(dB) = 80$
Answer:
$l(dB) = 80$
option 4

3 0

3 years ago

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Suppose a tank contains 400 gallons of salt water. If pure water flows into the tank at the rate of 7 gallons per minute and the

Strike441 [17]

Answer:

Step-by-step explanation:

This is a differential equation problem most easily solved with an exponential decay equation of the form

$y=Ce^{kt}$ . We know that the initial amount of salt in the tank is 28 pounds, so

C = 28. Now we just need to find k.

The concentration of salt changes as the pure water flows in and the salt water flows out. So the change in concentration, where y is the concentration of salt in the tank, is $\frac{dy}{dt}$ . Thus, the change in the concentration of salt is found in

$\frac{dy}{dt}=$ inflow of salt - outflow of salt

Pure water, what is flowing into the tank, has no salt in it at all; and since we don't know how much salt is leaving (our unknown, basically), the outflow at 3 gal/min is 3 times the amount of salt leaving out of the 400 gallons of salt water at time t:

$3(\frac{y}{400})$