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4 years ago

An Airbus A380 has 1 first class section 1 business class section and 4 economy class sections

Mathematics

1 answer:

Juliette [100K]4 years ago

7 0

Answer:

Yes, more than half are being used. (292 out of 546 used)

Step-by-step explanation:

The first part of this problem is to determine the total amount of seats on the plane so we can determine what half of the seats would be.

1st class has 14 seats.

Business-class has 64 seats.

And the 4 economy classes have 117 each, meaning 117 * 4 = 468

So the total amount of seats are 14 + 64 + 468 = 546

So half that amount would be 273.

Now we can determine the number of seats being used.

1st class has (6/7) * 14 = 12 seats used

Buisness-class has (7/16) * 64 = 28 seats being used

Economy has 4 * (7/13) * 117 = 4 * 63 = 252 seats being used

So the total amount of used seats would be 12 + 28 + 252 = 292.

Note that 292 is greater than 273, so there is indeed more than half the seats on the plane being used.

Cheers.

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For what value of x does 34x=27x-3

Anit [1.1K]

Answer:

Step-by-step explanation:

Let's solve your equation step-by-step.

34x=27x−3

Step 1: Subtract 27x from both sides.

34x−27x=27x−3−27x

7x=−3

Step 2: Divide both sides by 7.

7x\7=−3\7

x=−3/7

8 0

3 years ago

2. Write the equation of the line that passes

jenyasd209 [6]

Answer:

y = x + 1 which is marked by the broken line

6 0

3 years ago

Mohamed decided to track the number of leaves on the tree in his backyard each year The first year there were 500 leaves Each ye

svetlana [45]

Answer:

The required recursive formula is

$f(n)= 500\times(1.4)^{n-1}\\$

Step-by-step explanation:

Mohamed decided to track the number of leaves on the tree in his backyard each year.

The first year there were 500 leaves

$Year \: 1 = 500$

Each year thereafter the number of leaves was 40% more than the year before so that means

$Year \: 2 = 500(1+0.40) = 500\times 1.4\\$

For the third year the number of leaves increase 40% than the year before so that means

$Year \: 3 = 500\times 1.4(1+0.40) = 500 \times 1.4^{2}\\$

Similarly for fourth year,

$Year \: 4 = 500\times 1.4^{2}(1+0.40) = 500\times 1.4^{3}\\$

So we can clearly see the pattern here

Let f(n) be the number of leaves on the tree in Mohameds back yard in the nth year since he started tracking it then general recursive formula is

$f(n)= 500\times(1.4)^{n-1}\\$

This is the required recursive formula to find the number of leaves for the nth year.

Bonus:

Lets find out the number of leaves in the 10th year,

$f(10)= 500\times(1.4)^{10-1}\\\\f(10)= 500\times(1.4)^{9}\\\\f(10)= 500\times20.66\\\\f(10)= 10330$

So there will be 10330 leaves in the 10th year.

3 0

3 years ago

Read 2 more answers

Perform the indicated operation:

yaroslaw [1]

The expression $\frac{m^{2}-16\cdot n^{2}}{\frac{3\cdot m + 12\cdot n}{m\cdot n} }$ is equal to $\frac{m^{2}\cdot n}{3} -\frac{4\cdot m\cdot n^{2}}{3}$ .

<h3>How to simplify a composite algebraic expression</h3>

In this question we must use properties of <em>real</em> algebra to simplify a given expression as most as possible. We proceed to show each step with respective reason.

$\frac{m^{2}-16\cdot n^{2}}{\frac{3\cdot m + 12\cdot n}{m\cdot n} }$ Given
$\frac{m\cdot n \cdot (m^{2}-16\cdot n^{2})}{3\cdot m + 12\cdot n}$ $\frac{\frac{a}{b} }{\frac{c}{d} } = \frac{a\cdot d}{c\cdot b}$
$\frac{m\cdot n \cdot (m-4\cdot n)\cdot (m+4\cdot n)}{3\cdot (m+4\cdot n)}$ Distributive property/ $a^{2}-b^{2} = (a +b)\cdot (a-b)$
$\frac{m\cdot n \cdot (m-4\cdot n)}{3}$ Existence of the multiplicative inverse/Modulative property
$\frac{m^{2}\cdot n}{3} -\frac{4\cdot m\cdot n^{2}}{3}$ Distributive property/ $x^a\cdot x^{b} = x^{a+b}$ /Result