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4 years ago

11

Jim’s lunch account has a balance of $58.19. If lunch costs $2.74 per day, how many days will Jim be able to buy lunch before hi

s account runs out of money?

2 answers:

Tju [1.3M]4 years ago

7 0

21 days is how many days jim will be able to buy lunch before his account runs out of money

attashe74 [19]4 years ago

5 0

21 days until he runs out

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What is the solution to the inequality?<br><br> 4(2x+5)-2x more than or equal to 10

Alexxx [7]

See attached for the solution

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3 years ago

A half-filled cylindrical water tank has a water level of 20 feet high. The tank can hold 6000 cubic feet of water. Find the dia

Anettt [7]

Answer:

d = 13.8 feet

Step-by-step explanation:

Because we are talking about cubic feet of water, we need the formula for the VOLUME of a cylinder. That formula is

$V=\pi r^2h$

We will use 3.141592654 for pi; if the tank HALF filled with water is at 20 feet, then the height of the tank is 40 feet, so h = 40; and the volume it can hold in total is 6000 cubic feet. Filling in then gives us:

$6000=(3.141592654)(r^2)(40)$

Simplify on the right to get

$6000=125.6637061r^2$

Divide both sides by 125.6637061 to get that

$r^2=47.74648294$

Taking the square root of both sides gives you

r = 6.90988299

But the diameter is twice the radius, so multiply that r value by 2 to get that the diameter to the nearest tenth of a foot is 13.8

5 0

4 years ago

How do you solve this problem? 47 Fahrenheit =

mojhsa [17]

47 farenheit equals 29 celseus

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3 years ago

Read 2 more answers

If f(x) = 5x, what is f^-1(x)?<br> F(x) = -5x<br> of(x)-<br> F(x) = 5*<br> O F(x) = 5x

Oksana_A [137]

Answer:

f^-1 (x) = x/5

Step-by-step explanation:

f(x) = 5x

Replace f(x) with y

y= 5x

Exchange x and y

x = 5y

Solve for y

Divide each side by 5

x/5 = 5y/5

x/5 = y

The inverse is x/5

3 0

3 years ago

In a previous exercise we formulated a model for learning in the form of the differential equation dP dt = k(M − P) where P(t) m

GalinKa [24]

Answer:

$\frac{dP}{M-P}= kdt$

And we can integrate both sides of the equation using the following substitution:

$u= M-P, du =-dP$

And replacing we got:

$\int -\frac{du}{u} = kt +C$

$-ln (u)= kt+c$

If we multiply both sides by -1 we got:

$ln (u ) = -kt -c$

$ln (M-P) = -kt -c$

And using exponential in both sides of the equation we got:

$M-P = e^{-kt} e^{-c}$

And solving for P we got:

$P(t) = M- e^{-kt}e^{-c}$

And replacing $P_o =e^{-c}$ we got:

$P(t) = M - P_o e^{-kt}$

We can use the condition $P(0)=0$ and we got:

$0 = M -P_o e^0$

And we see that $M = P_o$ and replacing we got:

$P= M(1- e^{-kt})$

Step-by-step explanation:

For this case we aasume the following differential equation:

$\frac{dP}{dt}= k(M-P)$

Is a separable differential equation so we can do the following procedure:

$\frac{dP}{M-P}= kdt$

And we can integrate both sides of the equation using the following substitution:

$u= M-P, du =-dP$

And replacing we got:

$\int -\frac{du}{u} = kt +C$

$-ln (u)= kt+c$

If we multiply both sides by -1 we got:

$ln (u ) = -kt -c$

$ln (M-P) = -kt -c$

And using exponential in both sides of the equation we got:

$M-P = e^{-kt} e^{-c}$

And solving for P we got:

$P(t) = M- e^{-kt}e^{-c}$

And replacing $P_o =e^{-c}$ we got:

$P(t) = M - P_o e^{-kt}$

We can use the condition $P(0)=0$ and we got:

$0 = M -P_o e^0$

And we see that $M = P_o$ and replacing we got:

$P= M(1- e^{-kt})$

8 0

3 years ago