I NEED HELP AS SOON AS POSSIBLE!

Mathematics

1 answer:

Masja [62]3 years ago

7 0

Think of it this way:

You are trying to make a ratio to the months with thirty days and the ones who don't.

So,

there are 4 months a year with 30 days.

That means that there are 12-4= 8 months a year that are not 40 days.

It said the months to 30 days first in the problem.

4:8

Divide each side by 4, since both are divisible by 4.

1:2

For every one month of 30 days, there are 2 months that do not have 30 days.

1 to 2 is your answer.

I hope this helps!
~kaikers

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I really need help and if you give wrong answer or link then you are gonna be reported!

k0ka [10]

Answer:

hope it helps you.......

3 0

2 years ago

This is a 3 part 1 question each part on how to locate the plane and a small explanation report working out the recovery details

sesenic [268]

Third leg.

The crew flies at a speed of 560 mi/h in direction N-20°-E.

The wind has a speed of 35 mi/h and a direction S-10°-E.

We then can draw this as:

We have to add the two vectors to find the actual speed and direction.

We will start by adding the x-coordinate (W-E axis):

$\begin{gathered} x=560\cdot\sin (20\degree)+35\cdot\sin (10\degree) \\ x\approx560\cdot0.342+35\cdot0.174 \\ x\approx191.53+6.08 \\ x\approx197.61 \end{gathered}$

and the y-coordinate (S-N axis) is:

$\begin{gathered} y=560\cdot\cos (20\degree)-35\cdot\cos (10\degree) \\ y\approx560\cdot0.940-35\cdot0.985 \\ y\approx526.23-34.47 \\ y\approx491.76 \end{gathered}$

Then, the actual speed vector is v3=(197.61, 491.76).

The starting location for the third leg is R2=(216.66, 167.67) [taken from the previous answer].

Then, we have to calculate the displacement in 20 minutes using the actual speed vector.

We can calculate the movement in each of the axis. For the x-axis:

$\begin{gathered} R_{3x}=R_{2x}+v_{3x}\cdot t \\ R_{3x}=216.66+197.61\cdot\frac{1}{3} \\ R_{3x}=216.66+65.87 \\ R_{3x}=282.53 \end{gathered}$

NOTE: 20 minutes represents 1/3 of an hour.

We can do the same with the y-coordinate:

$\begin{gathered} R_{3y}=R_{2y}+v_{3y}\cdot t \\ R_{3y}=167.67+491.76\cdot\frac{1}{3} \\ R_{3y}=167.67+163.92 \\ R_{3y}=331.59 \end{gathered}$

The final position is R3 = (282.53, 331.59).

To find the distance from the origin and direction, we transform the cartesian coordinates of R3 into polar coordinates:

The distance can be calculated as if it was a right triangle:

$\begin{gathered} d^2=x^2+y^2_{} \\ d^2=282.53^2+331.59^2 \\ d^2=79823.20+109951.93 \\ d^2=189775.13 \\ d=\sqrt[]{189775.13} \\ d\approx435.63 \end{gathered}$

The angle, from E to N, can be calculated as:

$\begin{gathered} \tan (\alpha)=\frac{y}{x} \\ \tan (\alpha)=\frac{331.59}{282.53} \\ \tan (\alpha)\approx1.1736 \\ \alpha=\arctan (1.1736) \\ \alpha=49.56\degree \end{gathered}$

If we want to express it from N to E, we substract the angle from 90°: