Find the measure of each angle: Complementary angles with measures (5x)° and (4x−18)°.

Ksenya-84 [330]

Answer: im not sure

Step-by-step explanation:

4 0

2 years ago

As a student, you are able to earn extra money by assisting your neighbors with odd jobs. If you charged $10.25 an hour for your

olga nikolaevna [1]

Answer:

821.95

Step-by-step explanation:

5 0

2 years ago

Finish the polynomial by adding one number to make it a perfect square<br><br> x^2+18x+

dsp73

Answer:

81

Step-by-step explanation:

${x}^{2} + 18x + {(( \frac{1}{2})(18)) }^{2}$

${x}^{2} + 18x + {9}^{2}$

${x}^{2} + 18x + 81$

${(x + 9)}^{2}$

6 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°:

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

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

9 months ago

Solve the system of equations by graphing: -1/3x + y=-1, y=4+1/3x

Arada [10]

$-\frac{1}{3}x$ + y = -1 ⇒ y = $\frac{1}{3}x$ - 1

To graph this line, plot a point at the y-intercept (0, -1), than plot the next point using the rise over run from the slope $(\frac{1}{3})$ by counting up 1 and to the right 3 of the y-intercept. This gives you a second point of (3. 0). Draw a line through those two coordinates.

Answer: Plot (0, -1) and (3, 0) and draw a line through them.

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y = 4 + $\frac{1}{3}x$ ⇒ y = $\frac{1}{3}x$ + 4

Same as above. Plot the y-intercept (0, 4) and then use rise over run from the slope to plot (3, 5).

Answer: Plot (0, 4) and (3, 5) and draw a line through them.

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You should end up with two PARALLEL lines. Since the lines never intersect, there are no solutions to this system of equations.

Answer: No Solution

3 0

3 years ago