The worlds heaviest lobster weighted 44 pounds 6 ounces. What is the lobsters weight in ounces. Explain your thinking

A motorboat moves across a lake. It begins at 50km from shore after 9 minutes it is 14 km from shore. What’s it’s distance from

agasfer [191]

Answer:

List all of the solutions.

50 k m =

9 m i n u t e s =

14 k m =

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Which is the same as 10-2?<br> A) 0.0001 <br> B) 0.001 <br> C) 0.01 <br> D) 100

IceJOKER [234]

The correct answer is C. This is because
10^-2 = 1/10^2 which is 0.01

3 0

3 years ago

What is the product of the tens on 14 and 42

BartSMP [9]

Answer:

The product of the tens in these two numbers is 4

Step-by-step explanation:

TO find this, first we need to find the number in the tens place for each of them. The tens place is the second from the decimal point.

14

42

Now we take these two and multiply them together to get the product.

1 * 4 = 4

3 0

3 years ago

Read 2 more answers

Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 lite

zhenek [66]

Answer:

Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

Let the concentration of salt be a gram/L

Let the amount salt in the tank at any time t be Q(t).

$\frac{dQ}{dt} =\textrm {incoming rate - outgoing rate}$

Incoming rate = (a g/L)×(1 L/min)

=a g/min

The concentration of salt in the tank at any time t is = $\frac{Q(t)}{10}$ g/L

Outgoing rate = $(\frac{Q(t)}{10} g/L)(1 L/ min)$ $\frac{Q(t)}{10} g/min$

$\frac{dQ}{dt} = a- \frac{Q(t)}{10}$

$\Rightarrow \frac{dQ}{10a-Q(t)} =\frac{1}{10} dt$

Integrating both sides

$\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt$

$\Rightarrow -log|10a-Q(t)|=\frac{1}{10} t +c$ [ where c arbitrary constant]

Initial condition when t= 20 , Q(t)= 15 gram

$\Rightarrow -log|10a-15|=\frac{1}{10}\times 20 +c$

$\Rightarrow -log|10a-15|-2=c$

Therefore ,

$-log|10a-Q(t)|=\frac{1}{10} t -log|10a-15|-2$ .......(1)

In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0 in equation (1) we get

$- log|10a|= -log|10a-15| -2$

$\Rightarrow- log|10a|+log|10a-15|= -2$

$\Rightarrow log|\frac{10a-15}{10a}|= -2$

$\Rightarrow |\frac{10a-15}{10a}|=e ^{-2}$

$\Rightarrow 1-\frac{15}{10a} =e^{-2}$

$\Rightarrow \frac{15}{10a} =1-e^{-2}$

$\Rightarrow \frac{3}{2a} =1-e^{-2}$

$\Rightarrow2a= \frac{3}{1-e^{-2}}$

$\Rightarrow a = 1.73$

Therefore the concentration of salt in the incoming brine is 1.73 g/L

8 0

3 years ago

Ravi will take the 6:37 PM train to San Diego. The train is estimated to arrive in San Diego in 3 hours 45 minutes. What is the

Ahat [919]

So 6+3 is 9 right? So let's just say it's 9:37 with a 45 min ride. 60 minutes are in a hour. 60 - 37 = 23. We have 23 minutes left until 10:00. 45- 23 = 22. Therefore it'd be 10:22 p.m.

7 0

3 years ago