What is the real part of 4

Pani-rosa [81]

Answer:

hahqhqqh thank you for your time and consideration and I will be in the future of our games are based on the future of our games are based on the future of. our games are based on the future of.

3 0

3 years ago

The speed of the current in a river is 6 mph. A ferry operator who works that part of the river is looking to buy a new boat for

attashe74 [19]

Lest organize and label the information the problem is giving us first:
We know that the total time of the trip is 9 hours, so $t_{t}=9$ . We also know that the total distance of the trip is 22.5 miles, so $d_{t}=22.5$ . And we also know that the speed of the current is 6 mph, so $S_{c}=6$ .
Now, lets organize and label the things we don't know:
We don't know the speed of the boat in the lake, so $S_{l}=?$ . We don't know the time of the trip against the current, so $t_{ac}=?$ . We don't know the time of the trip with the current, so $t_{wc}=?$

The next thing we are going to do is set up equations to relate the things we don'k know with the things we actually know.
If the distance of the whole trip is 22.5 miles, the distance of the trip against the current is $\frac{22.5}{2}=11.25$ , so $d_{ac}=11.25$ . Similarly, the distance with the current is 11.25, so $d_{wc}=11.25$ .

Now lets use the equation $speed= \frac{distance}{time}$ to solve our problem:
Speed of the boat against the current:
$S_{l}-6= \frac{11.25}{t_{ac} }$ equation (1)
Speed of the boat with the current:
$S_{l}+6= \frac{11.25}{t_{wc}}$ equation (2)
Notice that we don't know the values of $t_{ac}$ and $t_{wc}$ , but we cant take advantage of the fact that the total time of the trip is 9 hours, so $9=t_{ac}+t_{wc}$ . Lets solve this equation for $t_{wc}$ :
$t_{wc}=9-t_{ac}$ equation (3)

Now we can replace equation (3) in equation (2) to express our equations with only tow variables:
$S_{l}+6= \frac{11.25}{9-t_{ac} }$ equation (4)
Next, lets solve for $t_{ac}$ in equation (4):
$S_{l}+6= \frac{11.25}{9-t_{ac}}$
$9-t_{ac}= \frac{11.25}{S_{l}+6}$
$-t_{ac}= \frac{11.25}{S_{l}+6} -9$
$-t_{ac}= \frac{11.25-9S_{l}-54}{S_{l}+6}$
$-t_{ac}= \frac{-42.75-9S_{l}}{S_{l}+6}$
$t_{ac}= \frac{42.75+9S_{l}}{S_{l}+6}$ equation (5)

Replace equation (5) in equation (1):
$S_{l}-6= \frac{11.25}{t_{ac}}$
$S_{l}-6= \frac{11.25}{ \frac{42.75+9S_{l}}{S_{l}+6} }$ equation (6)

Finally, we can solve equation (6) to find the speed of the boat in the lake:
$(S_{l}-6)(42.75+9S_{l})=11.25(S_{l}+6)$
$42.75S_{l}+9S_{l}^{2} -256.5-54S_{l}=11.25S_{l}+67.5$
$9S_{l}^{2}-11.25S_{l}-256.5=11.25S_{l}+67.5$
$9S_{l}^{2}-22.5S_{l}-324=0$
Using the quadratic formula to solve our quadratic equation, we get that S_{l}= 7.38 or $S_{l}=-4.88$ . Since speed cannot be negative, the solution of our equation is $S_{l}=7.38$ .

We can conclude that the speed of the boat on the lake <span>must be 7.38 mph in order for it to serve the ferry operator’s needs.</span>

8 0

3 years ago

I need help with this word problem:

elena-14-01-66 [18.8K]

Answer: 5 bikes and 3 tricycles

Step-by-step explanation: 10/2=5 bikes

9/3=3 tricycles

6 0

3 years ago

Simplify and solving practice 2

pentagon [3]

Answer:

The what

please but a picture please

5 0

3 years ago

katrin2010 [14]

Answer:

∆DPL ~ ∆QAZ

Step-by-step explanation:

In the diagram given, the following can be deduced:

<D is congruent to <Q

<P is congruent to <A

<L is congruent to <Z

From these, looking at the letter arrangement, the similarity statement can be written as:

∆DPL ~ ∆QAZ

4 0

3 years ago

What is the real part of 4 - 5i