1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Tamiku [17]
2 years ago
15

PLEASE HELP DUE SOON MORE POINTS IF CORRECT!!

Mathematics
2 answers:
SVETLANKA909090 [29]2 years ago
5 0

Answer:

y=4x+35

Step-by-step explanation:

yuradex [85]2 years ago
4 0
Okay I’m sorry but this is wownsukan enosksbeksn e skxoosnwbd
You might be interested in
An Olympic-sized swimming pool holds 2,500,000 liters of water. How many gallons of water does an Olympic-sized swimming pool ho
tankabanditka [31]

Answer:

The Olympic sized swimming pool hold 660501.9 gallons of water.

Step-by-step explanation:

Given that:

Water hold by an Olympic sized swimming pool = 2500000 liters

We know that;

1 gallon = 3.785 liters

Therefore,

We will divide the total number of water in liters by 3.785 to find the number of gallons.

Gallons of water in pool = \frac{2500000}{3.785}

Gallons of water in pool = 660501.9 gallons

Hence,

The Olympic sized swimming pool hold 660501.9 gallons of water.

5 0
3 years ago
A train is spotted 10 miles south and 8 miles west of an observer at 2:00 pm. At 3:00 pm the train is spotted 5 miles north and
kodGreya [7K]

Answer:

a. The distance the train travelled in the first hour is approximately 28.3 miles

b. The location of the train at 5:00 p.m. is 53 miles east, and 46 miles west

c. The location of the train at any given time by the function, f(t) = (-8 + 24·t, -10 + 15·t)

d. The train does not collide with the cyclist when the bike goes over the train tracks

Step-by-step explanation:

a. The given information on the train's motion are;

The location south the train is spotted = 10 miles south and 8 miles west

The time the observer spotted the train = 2:00 pm

The location the train is spotted at 3:00 p.m. = 5 miles north and 16 miles east

Therefore, the difference between the two times the train was spotted, t = 3:00 p.m. - 2:00 p.m. = 1 hour

Making use of the coordinate plane for the two locations the train was spotted, we have;

The initial location of the train = (-10, -8)

The final location of the train = (5, 16)

Therefore the distance the train travelled in the first hour is given by the formula for finding the distance, 'd', between two points, (x₁, y₁) and (x₂, y₂) as follows;

d = \sqrt{\left (x_{2}-x_{1}  \right )^{2}+\left (y_{2}-y_{1}  \right )^{2}}

Therefore;

d = \sqrt{\left (5-(-10)  \right )^{2}+\left (16-(-8)  \right )^{2}} = 3 \cdot\sqrt{89}

The distance the train travelled in the first hour, d = 3·√89 ≈ 28.3 miles

b. The speed of the train, v = (Distance travelled by the train)/Time

∴ v ≈ 28.3 miles/(1 hour) = 28.3 miles per hour

The speed of the train in the first hour, v ≈ 28.3 mph

The direction of the train, θ, is given by the arctangent of the slope, 'm', of the path of the train;

\therefore The  \  slope  \  of \ the \  path  \ of \  the \  train, \, m =tan(\theta) = \dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}

∴ m = tan(θ) = (5 - (-10))/(16 - (-8)) = 0.625

c. Distance = Velocity × Time

At 5:00 p.m., we have;

The time difference, Δt = 5:00 p.m. - 3:00 p.m. = 2 hours

The distance, d₁ = (28.3 mph × 2 hours = 56.6 miles

Using trigonometry, we have the horizonal distance travelled, 'Δx', in the 2 hours is given as follows;

Δx = d₁ × cos(θ)

∴ Δx = 56.6 × cos(arctan(0.625)) ≈ 48

The increase in the horizontal position of the train, relative to the point (5, 16), Δx ≈ 48 miles

The vertical distance increase in the two hours, Δy is given as follows;

Δy = 56.6 × sin(arctan(0.625)) ≈ 30

The increase in the vertical position of the train, relative to the point (5, 16), Δy ≈ 30 miles miles

Therefore; the location of the train at 5:00 p.m. = ((5 + 48), (16 + 30)) = (53, 46)

The location of the train at 5:00 p.m. = 53 miles east, and 46 miles west

c. The function, 'f', that would give the train's position at time-t is given as follows;

The P = f(28.3·t, θ)

Where;

28.3·t = √(x² + y²)

θ = arctan(y/x)

Parametric equations

y - 5 = 0.625·(x - 16)

∴ y = 0.625·x - 10 + 5

The equation of the train's track is therefore, presented as follows;

y = 0.625·x - 5

d = 28.3·t

The y-component of the velocity, v_y = 3*√89 mph × sin(arctan(0.625)) = 15 mph

Therefore, we have;

y = -10 + 15·t

The x-component of the velocity, vₓ = 3*√89 mph × cos(arctan(0.625)) = 24 mph

Therefore, we have;

x = -8 + 24·t

The location of the train at any given time, 't', f(t) = (-8 + 24·t, -10 + 15·t)

d. The speed of the cyclist next to the observer at 2:00 p.m., v = 10 mph

The distance of the cyclist from the track = The x-intercept = 5/0.625 = 8

The distance of the cyclist from the track = 8 miles

The time it would take the cyclist to react the track, t = 8 miles/10 mph = 0.8 hours

The location of the train in 0.8 hours, is f(0.8) = (-8 + 24×0.8, -10 + 15×0.8)

∴ f(0.8) = (11.2, 2)

At the time the cyclist is at the track along the east-west axis, at the point (8, 0), the train is at the point (11.2, 2) therefore, the train does not collide with the cyclist when the bike goes over the train tracks.

8 0
3 years ago
−3z−(−z−2) Simplify to create an equivalent expression.
netineya [11]
-3z - (-z - 2)

In order to find the expression equivalent to our given expression, we have to simplify our given expression :)

-3z + z + 2

**Remember : the "z" and "2" become a positive because a negative times a negative is always a positive!**

Now, add like terms.

(-3z + z) + 2

Simplify.

-2z + 2

Look at our given answer choices.

Our answer is c. -2z + 2

~Hope I helped!~
4 0
3 years ago
Read 2 more answers
Need help fast please
Ray Of Light [21]

Answer:

B

Step-by-step explanation:

5 0
3 years ago
Please help <br><br> Look at the picture
Trava [24]

Answer:

-3/2

Step-by-step explanation:

Find rise / run. It goes down 3 (so -3) and right 2 (so positive 2)

-3/2

5 0
3 years ago
Read 2 more answers
Other questions:
  • Grace wants to buy gumballs out of a vending machine in an airport. Two gumballs cost $0.80. If Grace uses exact change, in how
    6·1 answer
  • Create an equation with at least two grouping symbols for which<br> there is no solution
    5·1 answer
  • Can someone help please?? <br><br> I’lginsu edthvukcstghk
    6·1 answer
  • HELP ASAP PLZ RADIANS AND DEGREES
    10·1 answer
  • Can you please help with this question
    7·1 answer
  • 100 POINTS!! Use the relationship between the angles in the figure to answer the question.
    12·2 answers
  • Five more than 3/4 of a number C is 26 find the number
    11·2 answers
  • 210+698-812+1453-1300
    13·2 answers
  • Plz help will be marked BRAINLIEST!!<br><br><br> Yed
    7·2 answers
  • What is the length of the hypotenuse of the triangle when x=14
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!