What is the Equivalent of 62.5%

Circle A is inscribed in a quadrilateral. What is the perimeter of the quadrilateral?

faust18 [17]

AB + CD = BC + DA
Where A, B, C, and D are points on the quad, and AB is the length of the line A to B, etc. 

Since you know three sides, you can easily solve for the fourth.

6 0

3 years ago

A recipe requires 31 cup of milk for each 41 cup of water. How

avanturin [10]

Step-by-step explanation:

here,

31 cup of milk require 41 cup of water.

1 cup of milk require 41/31 cup of water.

so, 41/31 cup of water is required for 1 cup of milk.

hope u get it..

6 0

3 years ago

I next help on the polynomial

slamgirl [31]

Answer:

u just have to cross multiply im pretty sure thats how my math teacher show me,

Step-by-step explanation:

first you take the one x over to the other x and that will make it 2x then do the same thing with -10 u do -10 times 10 and then u get -100 and then the answer is 2x-100 and depending on your teach if he/she wants you to stop their then u can but if u have to got further then u would divide 2 on both sides and then that would be 0.02

this is how my teacher showed us how to do it and i hope this helps you a lot and if it dont im sorry i tried for you!!!

have a nice day Hun

8 0

3 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°: