Which expression is equivalent to this quotient 6/28x+4 divided by 12/35x+5

The buying rate and selling rate of a Australian dollar in a bank are rs 80.20and rs 81.40respectively how much Australian dolla

olya-2409 [2.1K]

Answer:

5000 Australian Dollars

Step-by-step explanation:

To find out how many Australian dollars need to be sold, we first need to find the profit of a single dollar sold.

We will be using the formula for profit, which is:

Profit = Total Revenue - Total Cost

Now we define the available variables.

Total Revenue = 81.40

Total Cost = 80.20

Profit = 81.40 - 80.20

Profit = rs 1.20/dollar

Now we have to find how many dollars we have to sell to get a profit of rs 6000.

We simply divide the amount of profit that we want to the price per dollar.

Total Profit = 6000

Profit per dollar = 1.20

This give us:

6000 / 1.20 = 5000 Australian Dollars.

4 0

3 years ago

|2/3-3/2|3^2-17/3 a)11/16 b)11/6 c)79/6 d)83/6

vfiekz [6]

Answer: The correct answer is: [B]:

____________________________________

→ " $\frac{11}{6}$ " .

____________________________________________

Explanation:

We are asked to solve: |2/3-3/2|3^2-17/3 ;

Let's rewrite this problem as:

____________________________________________

Simplify:

" | $\frac{2}{3} - \frac{3}{2}$ | * $3^{2} - \frac{17}{3}$ | " ;

____________________________________________

7 0

3 years ago

Three times a number increased by 8 is no more than the number 4 find the number

loris [4]

The answer is less than or equal to -4. Because -4+8=4

3 0

3 years ago

An example of a unti price youd find in germany-

mamaluj [8]

An example of unit price in germany is the electricity

6 0

2 years ago

What is a quick and easy way to remember explicit and recursive formulas?

Oliga [24]

I always found derivation to be helpful in remembering. Since this question is tagged as at the middle school level, I assume you've only learned about arithmetic and geometric sequences.

First, remember what these names mean. An arithmetic sequence is a sequence in which consecutive terms are increased by a fixed amount; in other words, it is an additive sequence. If $a_n$ is the $n$ th term in the sequence, then the next term $a_{n+1}$ is a fixed constant (the common difference $d$ ) added to the previous term. As a recursive formula, that's

$a_{n+1}=a_n+d$

This is the part that's probably easier for you to remember. The explicit formula is easily derived from this definition. Since $a_{n+1}=a_n+d$ , this means that $a_n=a_{n-1}+d$ , so you plug this into the recursive formula and end up with

$a_{n+1}=(a_{n-1}+d)+d=a_{n-1}+2d$

You can continue in this pattern, since every term in the sequence follows this rule:

$a_{n+1}=a_{n-1}+2d$
$a_{n+1}=(a_{n-2}+d)+2d$
$a_{n+1}=a_{n-2}+3d$
$a_{n+1}=(a_{n-3}+d)+3d$
$a_{n+1}=a_{n-3}+4d$

and so on. You start to notice a pattern: the subscript of the earlier term in the sequence (on the right side) and the coefficient of the common difference always add up to $n+1$ . You have, for example, $(n-2)+3=n+1$ in the third equation above.

Continuing this pattern, you can write the formula in terms of a known number in the sequence, typically the first one $a_1$ . In order for the pattern mentioned above to hold, you would end up with

$a_{n+1}=a_1+nd$

or, shifting the index by one so that the formula gives the $n$ th term explicitly,

$a_n=a_1+(n-1)d$

Now, geometric sequences behave similarly, but instead of changing additively, the terms of the sequence are scaled or changed multiplicatively. In other words, there is some fixed common ratio $r$ between terms that scales the next term in the sequence relative to the previous one. As a recursive formula,

$a_{n+1}=ra_n$

Well, since $a_n$ is just the term after $a_{n-1}$ scaled by $r$ , you can write

$a_{n+1}=r(ra_{n-1})=r^2a_{n-1}$

Doing this again and again, you'll see a similar pattern emerge:

$a_{n+1}=r^2a_{n-1}$
$a_{n+1}=r^2(ra_{n-2})$
$a_{n+1}=r^3a_{n-2}$
$a_{n+1}=r^3(ra_{n-3})$
$a_{n+1}=r^4a_{n-3}$

and so on. Notice that the subscript and the exponent of the common ratio both add up to $n+1$ . For instance, in the third equation, $3+(n-2)=n+1$ . Extrapolating from this, you can write the explicit rule in terms of the first number in the sequence:

$a_{n+1}=r^na_1$

or, to give the formula for $a_n$ explicitly,

$a_n=r^{n-1}a_1$

6 0

3 years ago