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insens350 [35]
3 years ago
11

Joe is preparing 18 hot dogs for his party. however, joe only has 12 hot dog buns. how many more hot dog buns does joe need? 6 7

5 8
Mathematics
2 answers:
BlackZzzverrR [31]3 years ago
7 0
Hello!
18 - 12 = 6
Joe still needs 6 hotdog buns.
:)
Alja [10]3 years ago
3 0

Answer:

6 buns

Step-by-step explanation:

Given : Joe is preparing 18 hot dogs for his party.

            He has 12 hot dog buns.

To Find:  how many more hot dog buns does Joe need?

Solution :

Joe is preparing hot dogs = 18

He has hot dog buns = 12

So, more hot dog buns required =18-12

                                                       =6

No. of hot dog buns required more = 6

Hence Joe need 6 more hot dog buns .

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A tank contains 1600 L of pure water. Solution that contains 0.04 kg of sugar per liter enters the tank at the rate 2 L/min, and
goldfiish [28.3K]

Let S(t) denote the amount of sugar in the tank at time t. Sugar flows in at a rate of

(0.04 kg/L) * (2 L/min) = 0.08 kg/min = 8/100 kg/min

and flows out at a rate of

(S(t)/1600 kg/L) * (2 L/min) = S(t)/800 kg/min

Then the net flow rate is governed by the differential equation

\dfrac{\mathrm dS(t)}{\mathrm dt}=\dfrac8{100}-\dfrac{S(t)}{800}

Solve for S(t):

\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{S(t)}{800}=\dfrac8{100}

e^{t/800}\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{e^{t/800}}{800}S(t)=\dfrac8{100}e^{t/800}

The left side is the derivative of a product:

\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}

Integrate both sides:

e^{t/800}S(t)=\displaystyle\frac8{100}\int e^{t/800}\,\mathrm dt

e^{t/800}S(t)=64e^{t/800}+C

S(t)=64+Ce^{-t/800}

There's no sugar in the water at the start, so (a) S(0) = 0, which gives

0=64+C\impleis C=-64

and so (b) the amount of sugar in the tank at time t is

S(t)=64\left(1-e^{-t/800}\right)

As t\to\infty, the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.

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d1i1m1o1n [39]

Answer:

136

Step-by-step explanation:

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