All categories

2 years ago

Please help

Mathematics

2 answers:

Orlov [11]2 years ago

6 0

1. 163/10.5 = 15.5 litres
2. 138.9 * 15.5 = 2,152 pence which is 21$
3.163/50 =3.2 hours

Tema [17]2 years ago

5 0

Answer:

The answers are:

15.523 liters of petrol will be used.
The cost of petrol will be: 2156.2 pence
The journey will take 3 hours and 16 minutes approximately

Step-by-step explanation:

Given:

Distance of Hotel and car = 163 miles

Distance travelled by car per liter = 10.5 miles

The number of liters will be obtained by dividing the total distance by distance travelled in one liter.

So,

$number\ of\ liters = \frac{163}{10.5} = 15.523\ liters$

2. Cost for petrol:

Cost of petrol per liter = 138.9 pence per liter

Then the total cost of petrol will be:

$Total\ cost = 138.9*15.523 = 2156.1447 => 2156.2\ pences$

3. Time taken for journey

Distance travelled in an hour = speed = 50 miles per hour.

The speed is given by the formula:

$speed = \frac{distance}{time}$

Putting the values

$50 = \frac{163}{t}\\t = \frac{163}{50}\\t = 3.26\ hours$

It will take: 3 hours and 0.26*60 = 15.6 minutes => 3 hours 16 minutes

Hence,

The answers are:

15.523 liters of petrol will be used.
The cost of petrol will be: 2156.2 pence
The journey will take 3 hours and 16 minutes approximately

You might be interested in

The use of mathematical methods to study the spread of contagious diseases goes back at least to some work by Daniel Bernoulli i

harina [27]

Answer:

$y(t) = y_o e^{\beta t}$

$x(t) = x_o e^{\frac{-\alpha y_o }{\beta }[e^{-\beta t} - 1] }$

$\lim_{t \to \infty} x(t) = x_oe^{\frac{-\alpha y_o}{\beta } }$

Step-by-step explanation:

From the question we are told that

$\frac{dy}{y} = -\beta dt$

Now integrating both sides

$ln y = \beta t + c$

Now taking the exponent of both sides

$y(t) = e^{\beta t + c}$

=> $y(t) = e^{\beta t} e^c$

Let $e^c = C$

$y(t) = C e^{\beta t}$

Now from the question we are told that

$y(0) = y_o$

Hence

$y(0) = y_o = Ce^{\beta * 0}$

=> $y_o = C$

$y(t) = y_o e^{\beta t}$

From the question we are told that

$\frac{dx}{dt} = -\alpha xy$

substituting for y

$\frac{dx}{dt} = - \alpha x(y_o e^{-\beta t })$

=> $\frac{dx}{x} = -\alpha y_oe^{-\beta t} dt$

Now integrating both sides

$lnx = \alpha \frac{y_o}{\beta } e^{-\beta t} + c$

Now taking the exponent of both sides

$x(t) = e^{\alpha \frac{y_o}{\beta } e^{-\beta t} + c}$

=> $x(t) = e^{\alpha \frac{y_o}{\beta } e^{-\beta t} } e^c$

Let $e^c = A$

=> $x(t) =K e^{\alpha \frac{y_o}{\beta } e^{-\beta t} }$

Now from the question we are told that

$x(0) = x_o$

$x(0)=x_o =K e^{\alpha \frac{y_o}{\beta } e^{-\beta * 0} }$

=> $x_o = K e^{\frac {\alpha y_o }{\beta } }$

divide both side by $(K * x_o)$

=> $K = x_o e^{\frac {\alpha y_o }{\beta } }$

$x(t) =x_o e^{\frac {-\alpha y_o }{\beta } } * e^{\alpha \frac{y_o}{\beta } e^{-\beta t} }$

=> $x(t)= x_o e^{\frac{-\alpha * y_o }{\beta} + \frac{\alpha y_o}{\beta } e^{-\beta t} }$

=> $x(t) = x_o e^{\frac{\alpha y_o }{\beta }[e^{-\beta t} - 1] }$

Generally as t tends to infinity , $e^{- \beta t}$ tends to zero

$\lim_{t \to \infty} x(t) = x_oe^{\frac{-\alpha y_o}{\beta } }$

5 0

3 years ago

What is the plan for the set of stacked cubes?

Nadusha1986 [10]

$\large \gray { \boxed{{ \colorbox{g}{answer}}}}$

it only depends on the cubes in the pic

3 0

2 years ago

Who can help me with this number 5 I need help with

Ann [662]

Answer:

A.Y

B.X

C.They become opposite

Step-by-step explanation: A. When you reflect the x - axis you don't change the y - axis B: Same as A C: You multiply them both by -1 basicly

6 0

3 years ago

Help I will be marking brainliest!!!

Keith_Richards [23]

<h3>Answer: A. 18*sqrt(3)</h3>

=============================================

Explanation:

We'll need the tangent rule

tan(angle) = opposite/adjacent

tan(R) = TH/HR

tan(30) = TH/54

sqrt(3)/3 = TH/54 ... use the unit circle

54*sqrt(3)/3 = TH .... multiply both sides by 54

(54/3)*sqrt(3) = TH

18*sqrt(3) = TH

TH = 18*sqrt(3) which points to choice A as the final answer

----------------------

An alternative method:

Triangle THR is a 30-60-90 triangle.

Let x be the measure of side TH. This side is opposite the smallest angle R = 30, so we consider this the short leg.

The hypotenuse is twice as long as x, so TR = 2x. This only applies to 30-60-90 triangles.

Now use the pythagorean theorem

a^2 + b^2 = c^2

(TH)^2 + (HR)^2 = (TR)^2

(x)^2 + (54)^2 = (2x)^2

x^2 + 2916 = 4x^2

2916 = 4x^2 - x^2

3x^2 = 2916

x^2 = 2916/3

x^2 = 972

x = sqrt(972)

x = sqrt(324*3)

x = sqrt(324)*sqrt(3)

x = 18*sqrt(3) which is the length of TH.

A slightly similar idea is to use the fact that if y is the long leg and x is the short leg, then y = x*sqrt(3). Plug in y = 54 and isolate x and you should get x = 18*sqrt(3). Again, this trick only works for 30-60-90 triangles.

4 0

2 years ago

Please help I’ll give brainliest

AlekseyPX

It’s the 2nd on I’m pretty sure because it says decimal

8 0

3 years ago

Read 2 more answers