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ra1l [238]
2 years ago
15

Can someone please help me on this

Mathematics
1 answer:
Tanzania [10]2 years ago
7 0

a=2(-2)-1 = -4-1 =-5\\\\b=2(-1)-1 = -2 - 1 = -3\\\\c=2(1)-1 = 2-1 = 1\\\\d=2(2) -1 = 4 -1  =3

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What is the constant rate of change of (5,80) (4,64) and (6,96)
sveticcg [70]

Answer:

16

Step-by-step explanation:

The existence of the constant rate of change is given the ratio of y to x is the same. Then:

k_{1} = \frac{80}{5}

k_{1} = 16

k_{2} = \frac{64}{4}

k_{2} = 16

k_{3} = \frac{96}{6}

k_{3} = 16

In consequence, the constant rate of change is 16.

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2 years ago
A man has 1020304 eggs some of the eggs were spoiled he divided remaining eggs equally among 348 people how many eggs were spoil
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Answer:

316 eggs were spoiled.

Step-by-step explanation:

1020304 / 348 = 2931 remainder 316.

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What are the zeros of the quadratic function <br> y = 6x2 + 3x - 45
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2 years ago
Read 2 more answers
a nature clubs has 37 member.each member receives 15 fish to put aquarium . if 20 of total number of fish or put into a fish bow
PilotLPTM [1.2K]

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5 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
10 months ago
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