Ask question

Ask question

All categories

3 years ago

13

It takes a boat 2 hours to travel 22 miles upstream against a current. If the speed of the boat in still water is 17 miles per h

our, what is the speed of the current? The speed of the current is miles per hour.

1 answer:

lesya [120]3 years ago

4 0

Answer: I think about 5 or 6

Step-by-step explanation:

You might be interested in

What is the ansser <br><br><br> plase help

aniked [119]

Answer:

sir this isnt math

Step-by-step explanation:

put it in spanish next time

5 0

2 years ago

Read 2 more answers

What fracción is 20% of 40

lions [1.4K]

20 percent of 40 is 8. As a fraction that would be 8/1.

Hope this helps

6 0

3 years ago

Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y

irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

$y'(x)= x^2y(x)-\frac12y^2(x)$

Putting the value of y'(x) in equation (1)

$y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))$

Substituting x =0 and h= 0.2

$y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]$

$\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]$ [∵ y(0) =3 ]

$\Rightarrow y(0.2)\approx 2.7$

Substituting x =0.2 and h= 0.2

$y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]$

$\Rightarrow y(0.4)\approx 2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]$

$\Rightarrow y(0.4)\approx 1.9926$

Substituting x =0.4 and h= 0.2

$y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]$

$\Rightarrow y(0.6)\approx 1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]$

$\Rightarrow y(0.6)\approx 1.6593$

Substituting x =0.6 and h= 0.2

$y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]$

$\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]$

$\Rightarrow y(0.6)\approx 0.8800$

Substituting x =0.8 and h= 0.2

$y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]$

$\Rightarrow y(1.0)\approx 0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]$

$\Rightarrow y(1.0)\approx 0.9152$

Therefore the value of y(1)= 0.9152.

4 0

3 years ago

Here are the pictures I need help with (Math)

astraxan [27]

Idk I just want points can u give breanlyist please

3 0

3 years ago

Read 2 more answers

A the zoo, the zookeeper make 4,000 pounds of ice. The zookeeper can make 400 pounds per day. How many days will it take to make

Nataliya [291]

Answer:

10 days.

Step-by-step explanation:

to find what you multiply from 400 to get to 4000, you divide 4000 by 400, which is 10, plug that in, 400 pounds x 10 days = 4000 pounds.

6 0

2 years ago

Read 2 more answers