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alexandr1967 [171]
2 years ago
10

PLEASE HELP FOR EACH OF THE FIGURES, WRITE AN ABSOLUTE VALUE EQUATION

Mathematics
1 answer:
Likurg_2 [28]2 years ago
5 0

The solution set 0 and 8 are the true values of the absolute value equation

The absolute value equation that has the a solution set of 0 and 6 is |x - 3| = 3

<h3>How to determine the absolute value equation?</h3>

From the figure, the solution sets of the absolute value equation are given as:

x = {0, 6}

Calculate the mean of the solution set

Mean = 0.5 * (0 + 6)

Mean = 3

Calculate the difference of the solution set divided by 2

Difference = (6 - 0)/2

Difference = 3

The absolute value equation is the represented as:

|x - Mean| = Difference

Substitute known values

|x - 3| = 3

Hence, the absolute value equation that has the a solution set of 0 and 6 is |x - 3| = 3

Read more about absolute value equation at:

brainly.com/question/2166748

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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°:

\beta=90\degree-\alpha=90-49.56=40.44\degree

Answer: the final location can be represented with the vector (282.53, 331.59).

1) The distance from the origin is 435.63 miles and

2) the direction is N-40°-E.

7 0
1 year ago
How do I find the variable?<br><br> N x 1/5 = 2/15<br><br> N / 1/5 = 1/3
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First you need to get all variables on one side. You do that by multiplying the reciprocal of the fraction to everything.
N(5/1) x 1/5(5/1)=2/15(5/1)
5n=2/15 x 5/1
Then solve the multiplication problem
5n= 10/15
Then you should reduce the fraction
5n=2/3
Then divide both sides by 5
5n/5=N
2/3 divided by 5 =2/15

N=2/15




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