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3 years ago

I need help asap :)/)/)

Mathematics

1 answer:

liraira [26]3 years ago

6 0

Answer:

the first one

Step-by-step explanation:

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What is the greatest common factor (GCF) of 48 and 56? A. 168 B. 8 C. 4 D. 336

Sergio039 [100]

Answer:

B. 8

Step-by-step explanation:

48/8=6

56/8=7

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3 years ago

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Each member of the swim team must pay $76 to join the team. There are 376 members. How much does the team receive in dues?

Basile [38]

Answer:

$28,576

Step-by-step explanation:

$76 (amount that each member pays) * 376 (total members) = $28,576 (amount that will be received in dues)

I hope this helps :)

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3 years ago

Find the slope of the line whose equation is 0.05x - 0.03y = 9.

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1.66666666

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3 years ago

A 100 gallon tank initially contains 100 gallons of sugar water at a concentration of 0.25 pounds of sugar per gallon suppose th

Vsevolod [243]

At the start, the tank contains

(0.25 lb/gal) * (100 gal) = 25 lb

of sugar. Let $S(t)$ be the amount of sugar in the tank at time $t$ . Then $S(0)=25$ .

Sugar is added to the tank at a rate of P lb/min, and removed at a rate of

$\left(1\frac{\rm gal}{\rm min}\right)\left(\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm gal}\right)=\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm min}$

and so the amount of sugar in the tank changes at a net rate according to the separable differential equation,

$\dfrac{\mathrm dS}{\mathrm dt}=P-\dfrac S{100}$

Separate variables, integrate, and solve for S.

$\dfrac{\mathrm dS}{P-\frac S{100}}=\mathrm dt$

$\displaystyle\int\dfrac{\mathrm dS}{P-\frac S{100}}=\int\mathrm dt$

$-100\ln\left|P-\dfrac S{100}\right|=t+C$

$\ln\left|P-\dfrac S{100}\right|=-100t-100C=C-100t$

$P-\dfrac S{100}=e^{C-100t}=e^Ce^{-100t}=Ce^{-100t}$

$\dfrac S{100}=P-Ce^{-100t}$

$S(t)=100P-100Ce^{-100t}=100P-Ce^{-100t}$

Use the initial value to solve for C :

$S(0)=25\implies 25=100P-C\implies C=100P-25$

$\implies S(t)=100P-(100P-25)e^{-100t}$

The solution is being drained at a constant rate of 1 gal/min; there will be 5 gal of solution remaining after time

$1000\,\mathrm{gal}+\left(-1\dfrac{\rm gal}{\rm min}\right)t=5\,\mathrm{gal}\implies t=995\,\mathrm{min}$

has passed. At this time, we want the tank to contain

(0.5 lb/gal) * (5 gal) = 2.5 lb

of sugar, so we pick P such that

$S(995)=100P-(100P-25)e^{-99,500}=2.5\implies\boxed{P\approx0.025}$

5 0

3 years ago