Brandon buys a radio for $44.83 in a state where the sales tax is 7%.

RSB [31]

8/14 cause 4+4 is 8 7+7 14

4 0

3 years ago

Suppose an unknown radioactive substance produces 8000 counts per minute on a Geiger counter at a certain time, and only 500 cou

mariarad [96]

Answer:

The half-life of the radioactive substance is of 3.25 days.

Step-by-step explanation:

The amount of radioactive substance is proportional to the number of counts per minute:

This means that the amount is given by the following differential equation:

$\frac{dQ}{dt} = -kQ$

In which k is the decay rate.

The solution is:

$Q(t) = Q(0)e^{-kt}$

In which Q(0) is the initial amount:

8000 counts per minute on a Geiger counter at a certain time

This means that $Q(0) = 8000$

500 counts per minute 13 days later.

This means that $Q(13) = 500$ . We use this to find k.

$Q(t) = Q(0)e^{-kt}$

$500 = 8000e^{-13k}$

$e^{-13k} = \frac{500}{8000}$

$\ln{e^{-13k}} = \ln{\frac{500}{8000}}$

$-13k = \ln{\frac{500}{8000}}$

$k = -\frac{\ln{\frac{500}{8000}}}{13}$

$k = 0.2133$

So

$Q(t) = Q(0)e^{-0.2133t}$

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

$Q(t) = Q(0)e^{-0.2133t}$

$0.5Q(0) = Q(0)e^{-0.2133t}$

$e^{-0.2133t} = 0.5$

$\ln{e^{-0.2133t}} = \ln{0.5}$

$-0.2133t = \ln{0.5}$

$t = -\frac{\ln{0.5}}{0.2133}$

$t = 3.25$

The half-life of the radioactive substance is of 3.25 days.

7 0

3 years ago

The number 900 has an interesting property in that the only primes that divide into it evenly are 2, 3, and 5. What are the next

Tasya [4]

900 = 2 · 2 · 3 · 3 · 5 · 5

Next numbers:

2 · 2 · 2 · 3 · 3 · 3 · 5 · 5 · 5 = 900 · 15 = 13,500

2 · 2 · 2 · 2 · 3 · 3 · 3 · 3 · 5 · 5 · 5 · 5 = 13,500 · 15 = 202,500

202,500 · 15 = 3,037,500

3,037,500 · 15 = 45,562,500

....

3 0

4 years ago

Kate baked a berry pie. The recipe called for 7 ounces of blackberries, 7 ounces of raspberries, and 6 ounces of strawberries. I

saul85 [17]

Answer:

Step-by-step explanation:

The recipe for baking a berry pie called for 7 ounces of blackberries, 7 ounces of raspberries, and 6 ounces of strawberries.

Since she only found 12-ounce containers of berries, she bought 1 container of each kind of berry.

After she baked the pie,

The number of ounces of blackberries left would be 12 - 7 = 5 ounces.

The number of ounces of raspberries left would be 12 - 7 = 5 ounces.

The number of ounces of strawberries left would be 12 - 6 = 6 ounces.

Total number of ounces of berries that Kate has left would be

5 + 5 + 6 = 16 berries

3 0

3 years ago

Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.

natulia [17]

Answer:

$g'(x)=\frac{2e^x-e^{-\frac{x}{2}}}{2(e^x+e^{-\frac{x}{2}})}$

Step-by-step explanation:

We are given that a function

$g(x)=ln(e^x+e^{-\frac{x}{2}})$

We have to find the derivative of function

Differentiate w.r.t x

$g'(x)=\frac{1}{e^x+e^{-\frac{x}{2}}}\times (e^x+e^{-\frac{x}{2}}\times (-\frac{1}{2}))$

By using formula

$\frac{d(lnx)}{dx}=\frac{1}{x}$

$\frac{d e^x}{dx}=e^x$

$g'(x)=\frac{e^x-\frac{e^{-\frac{x}{2}}}{2}}{e^x+e^{-\frac{x}{2}}}$

Hence, the derivative function

$g'(x)=\frac{2e^x-e^{-\frac{x}{2}}}{2(e^x+e^{-\frac{x}{2}})}$

3 0

4 years ago