Select the phrase that best completes the sentence.

Dan had a large piece of wood that was 10 feet

Doss [256]

Answer:

40 pieces of wood

Step-by-step explanation:

Find how many pieces he cut by dividing 10 by 1/4

10 / (1/4)

= 40

So, Dan cut 40 pieces of wood

7 0

3 years ago

V/4 + 7 = 8<br><br> (also if you could maybe give an explanation id appreciate it a lot.)

Rufina [12.5K]

Answer:

Step-by-step explanation:

1. Subtract 7 on both sides

2. multiply by 4 on both sides

3. v = 4

6 0

3 years ago

Read 2 more answers

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

Find out distinct 3 letter word that can be formed from using the word singapore

Greeley [361]

Answer:

Sing

Gap

Nape

There are more than 3 distinct words

3 0

3 years ago

The initial value of a quantity Q (at year t = 0) is 112.8 and the quantity is decreasing by 23.4% per year. a) Write a formula

Svetach [21]

Answer:

a) $Q(t) = 112.8*(0.766)^{t}$

b) When t = 10, Q = 7.845.

Step-by-step explanation:

The value of a quantity after t years is given by the following formula:

$Q(t) = Q_{0}(1 + r)^{t}$

In which $Q_{0}$ is the initial quantity and r is the rate that it changes. If it increases, r is positive. If it decreases, r is negative.

a) Write a formula for Q as a function of t.

The initial value of a quantity Q (at year t = 0) is 112.8.

This means that $Q_{0} = 112.8$ .

The quantity is decreasing by 23.4% per year.

This means that $r = -0.234$

So

$Q(t) = 112.8*(1 - 0.234)^{t}$

$Q(t) = 112.8*(0.766)^{t}$

b) What is the value of Q when t = 10?

This is Q(10).

$Q(t) = 112.8*(0.766)^{t}$

$Q(t) = 112.8*(0.766)^{10} = 7.845$

When t = 10, Q = 7.845.

4 0

4 years ago