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shusha [124]
3 years ago
5

James covers a distance of 8 meters in the first second of a 100 meter race. Then, he sprints at a speed of 9 meters per second.

How far is James from the finish line 9 seconds after the race has started? a. 11 c. 28 b. 20 d. 80
Mathematics
2 answers:
AveGali [126]3 years ago
8 0
So James covers 8 meters in the first second....then he sprints 9 meters per second......
9 seconds after the race....so there are 8 seconds that he is sprinting 9 meters per second.....(8 * 9) = 72 meters....plus the 8 meters for the 1st second = (72 + 8) = 80 meters....
and if it is 100 meter race, and he has ran 80 meters, then that leaves 20 meters from the finish line....answer is D
vichka [17]3 years ago
5 0

Answer:

Option b. 20 meters

Step-by-step explanation:

James covers a distance of 8 meters in the first second of 100 meter race.

Then, he sprints at a speed of 9 meters per second.

We have to find how far is James from the finish line 9 seconds after the start of the race.

James covers distance in first second = 8 meters.

Time remaining in 9 seconds after running 8 meter = 9 - 1 =  8 seconds.

In another 8 seconds James covers the distance = speed × time

 = 9 × 8 = 72 meters

Total distance covered in 9 seconds = 8 + 72 = 80  meters

Distance remaining in 100 meter race after 9 seconds = 100 - 80 = 20 meters.

James is 20 meters far from the finish line 9 seconds after the race has started.

Option b. 20 meters is the answer.

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Svetradugi [14.3K]

Answer:

x = - 23

Step-by-step explanation:

Given

- x - 7 = 16 ( Isolate - x by adding 7 to both sides )

- x = 23 ( multiply both sides by - 1 )

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4 0
3 years ago
A ship leaves port at 10 miles per hour, with a heading of N 35° W. There is a warning buoy located 5 miles directly north of th
Leno4ka [110]

The value of the angle subtended by the distance of the buoy from the

port is given by sine and cosine rule.

  • The bearing of the buoy from the is approximately <u>307.35°</u>

Reasons:

Location from which the ship sails = Port

The speed of the ship = 10 mph

Direction of the ship = N35°W

Location of the warning buoy = 5 miles north of the port

Required: The bearing of the warning buoy from the ship after 7.5 hours.

Solution:

The distance travelled by the ship = 7.5 hours × 10 mph = 75 miles

By cosine rule, we have;

a² = b² + c² - 2·b·c·cos(A)

Where;

a = The distance between the ship and the buoy

b = The distance between the ship and the port = 75 miles

c = The distance between the buoy and the port = 5 miles

Angle ∠A = The angle between the ship and the buoy = The bearing of the ship = 35°

Which gives;

a = √(75² + 5² - 2 × 75 × 5 × cos(35°))

By sine rule, we have;

\displaystyle \frac{a}{sin(A)} = \mathbf{ \frac{b}{sin(B)}}

Therefore;

\displaystyle sin(B)= \frac{b \cdot sin(A)}{a}

Which gives;

\displaystyle sin(B) = \mathbf{\frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }}

\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx 37.32^{\circ}

Similarly, we can get;

\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx \mathbf{ 142.68^{\circ}}

The angle subtended by the distance of the buoy from the port, <em>C</em> is therefore;

C ≈ 180° - 142.68° - 35° ≈ 2.32°

By alternate interior angles, we have;

The bearing of the warning buoy as seen from the ship is therefore;

Bearing of buoy ≈ 270° + 35° + 2.32° ≈ <u>307.35°</u>

Learn more about bearing in mathematics here:

brainly.com/question/23427938

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Mazyrski [523]
Answer:

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240÷4 is 60 cupcakes for the party
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4 years ago
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