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Kobotan [32]
3 years ago
15

A vector is used to express a car's change in position, or displacement, by tracking its motion over a large area defined by a c

oordinate grid. If the car begins at (−3,−5) and ends at (5,9), which of these expresses the car's displacement in vector form?
A
(14,8)

B
(10,12)

C
(8,14)

D
(2,4)
Mathematics
1 answer:
Mnenie [13.5K]3 years ago
7 0

Answer:

The correct answer is option C: (8, 14).

Step-by-step explanation:

The car's displacement in vector form can be found by subtracting the initial points from the final points.    

Initial: (x₁, y₁) = (-3, -5)

Final: (x₂, y₂) = (5, 9)          

For the x-coordinate, we have:  

d_{x} = x_{2} - x_{1} = 5 - (-3) = 8

And for the y-coordinate, we have:

d_{y} = y_{2} - y_{1} = 9 - (-5) = 14

The car's displacement in vector form is:

d = (d_{x}, d_{y}) = (8, 14)  

Therefore, the correct answer is option C: (8, 14).

I hope it helps you!                        

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an object undergoes an acceleration of 8 metre per second square starting from rest find the distance travelled in one second​
garik1379 [7]

Answer:

4m

Step-by-step explanation:

Initial velocity (u) = 0,

Uniform acceleration (a) = 8 m/s²

Time (t) = 1 s

Distance travelled in 1 second

= ut + 1/2at²

= 0 + 1/2× 8 ×1²

= 4 m  Ans.

7 0
3 years ago
PLS HELPP QUICK QUESTION AND OPTIONS ARE BELOW IN ATTACHED FILE....
podryga [215]
Altogether there are 14 flavours of oatmeal.
The number of grams of fiber (per serving) is shown in the list/table.

What you're expected to do is to count the number of flavours that has the given number of grams of fibre (0-1, 2-3,...) and drag the appropriate tile to the appropriate location in the graph.

For example, for 0-1 grams of fibre.  I see there is are 2- "1", and no zeroes.  So there is only one single flavour that has 0-1 gram of fibre in a serving.  You will need to drag the tile marked "2" in the first position (0-1).

For 2-3 grams, I see 3 "2" and 3 "3" for a total of six flavours.  So you would drag the tile marked "6" to the second position in the graph (2-3).

Similarly, for 4-5 grams, I see 4-"4" and 1-"5", so again you drag the tile marked "5" to the third position on the graph.

Do the same for 6-7 grams, I see only 1-"6", so drag the tile marked "1" to the fourth position, then you're done!

Check: 2+6+5+1=14  ✔
3 0
4 years ago
A total of 560 tickets were sold
sweet-ann [11.9K]

Answer:

Student.... 420

Adult.......140

6 0
3 years ago
What is the approximate circumference of the circle shown below
erma4kov [3.2K]

Answer: A) 46.5 cm

Step-by-step explanation:

Circumference = 2 π r or Circumference = π d

In this case, they give us the diameter = 14.8 cm. Therefore the radius = 14.8/2 = 7.4 cm.

π ≈ 3.14.

We can use either equation above, but because the diameter was given, we will use c = πd for simplicity.

C = π d

= (3.14)(14.8)

= 46.47 cm, which rounds to 46.5 cm

5 0
3 years ago
A ship embarked on a long voyage. At the start of the voyage, there were 300 ants in the cargo hold of the ship. One week into t
Pie

Answer:

It takes 1 week for the ant population to double

It takes 1.58 weeks for the ant population to triple

It takes 5.06 weeks for the ant population to reach 10000

It takes 5.09 weeks for the ant population to be 200 times the ant eater population.

Step-by-step explanation:

It takes only 1 week for the population to double from 300 to 600

We can model the population of ant (or anteater) as the following:

p = ae^{kt}

Where a = 300 is the initial population at t = 0

When t = 1, P = 600

600 = 300e^{1k}

e^k = 2

k = ln2 = 0.693

When the population tripled, p/a = 3

e^{kt} = 3

kt = ln 3 = 1.1

t = 1.1/k = 1.1/0.693 = 1.58 weeks.

When there are 10000 ants on board, p = 10000:

300e^{kt} = 10000

e^{kt} = 10000/300 = 33.33

kt = ln33.33 = 3.51

t = 3.51 / k = 3.51 / 0.693 = 5.06 weeks.

Similarly for anteater, at t = 0 there are 17 of them so A = 17. We can solve for their K parameter if the population doubled after 3.2 weeks

e^{3.2K} = P/A = 2

3.2K = ln2

K = ln2/3.2 = 0.2166

At the time there are 200 ants per anteater

p = 200P

300e^{kt} = 200*17e^{Kt}

e^{kt - Kt} = 200*17/300

e^{0.693t - 0.2166t} = 11.33

e^{0.4765t} = 11.33

0.4765t = ln11.33

t = ln11.33/0.4765 = 5.09 weeks

6 0
3 years ago
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