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Studentka2010
12 days ago
5

In year 3 it is expected that the total value of clothing sales will reach 32 million if the total value of ASCO sells Remains t

he Same as year 2what percentage of sales clothing account for in year three
Mathematics
1 answer:
tester [3.9K]12 days ago
3 0

Answer:

The sales figure for the third year is [\frac{32\ \text{mn}-x\ \text{mn}}{x\ \text{mn}}\times 100\%].

Step-by-step explanation:

Since sales data for the second year is unavailable, we denote it as x million.

Year 3's total sales amount to 32 million.

Calculate the sales account for year 3 as follows:

\text{Percentage of sales for year 3}=\frac{\text{Total Sales in year 3}-\text{Total Sales in year 2}}{\text{Total Sales in year 2}}\times 100\%

                                                =\frac{32\ \text{mn}-x\ \text{mn}}{x\ \text{mn}}\times 100\%

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Choose the solutions of the following systems of equations x^2+y^2=6 and x^2-y=6
AnnZ [3877]

Answer:

The solutions are (-√6,0) and (√6,0)

(-√5,-1) and (√5,-1)

Step-by-step explanation:

We start with the equations:

{x}^{2} + {y}^{2} = 6

and

{x}^{2} - y = 6

Rearranging the second equation to make y the subject results in;

y = {x}^{2} - 6

Substituting into the first equation gives us:

{x}^{2} + ({x}^{2} - 6 )^2= 6

x = \pm \sqrt{6}

and x = \pm \sqrt{5}

The solutions remain as (-√6,0) and (√6,0)

(-√5,-1) and (√5,-1)

7 0
7 days ago
A rectangular schoolyard is to be fenced in using the wall of the school for one side and 150
lawyer [4008]

Answer:

Part 1) The equation is A(x)=150x-2x^2

Part 2) When x=40 m, the area of the schoolyard is A=2,800 m^2

Part 3) The valid domain consists of all real numbers exceeding zero and below 75 meters

Step-by-step explanation:

Part 1) Formulate an expression for A(x)

Let

x -----> the length of the rectangular school yard

y ---> the width of the rectangular school yard

It is known that

The perimeter for the fencing (taking the school wall as one side) is

P=2x+y

P=150\ m

thus

150=2x+y

y=150-2x -----> this is equation A

The area of the rectangular school yard is

A=xy ----> this is equation B

Substituting equation A into equation B yields

A=x(150-2x)

A=150x-2x^2

Change to function notation

A(x)=150x-2x^2

Part 2) What is the area when x=40?

With x equal to 40 m

substitute the value into the expression from Part 1 to determine A

A(40)=150(40)-2(40^2)

A(40)=2,800\ m^2    

Part 3) What would be a suitable domain for A(x) in this scenario?

We understand that

A signifies the area of the rectangular school yard

x characterizes the length of the rectangular school yard

It follows that

A(x)=150x-2x^2

This forms a vertical parabola opening downwards

The vertex indicates a maximum point

The x-coordinate of the vertex corresponds to the length that maximizes the area

The y-coordinate of the vertex denotes the maximum area

The vertex corresponds to (37.5, 2812.5)

Refer to the accompanying figure

Consequently,

The peak area achieved is 2,812.5 m^2

The x-intercepts are located at x=0 m and x=75 m

The domain for A is the range -----> (0, 75)

All real numbers greater than zero and less than 75 meters

5 0
11 days ago
Need help on 4.9.2 project: performance task: the subway stop
tester [3916]
You might have better success by searching for answers individually:)
3 0
6 days ago
If Ana devotes all her time to making fudge, she can make 3 pounds of fudge an hour, and if he devotes all her time to making to
lawyer [4008]

Answer: Ana should produce more fudge, and Leo should produce more toffee.

Step-by-step explanation:

Comparative advantage defined: A person or country has a comparative advantage producing a good if their opportunity cost for it is lower than someone else's.

In this case, Ana and Leo will both benefit if they focus on the product for which they have the lower opportunity cost.

(a) Ana’s production possibilities:

If she spends all her time making fudge: 3 pounds; or toffee: 2 pounds.

Opportunity cost of 1 pound fudge = 2/3 = 0.66 pounds toffee

Opportunity cost of 1 pound toffee = 3/2 = 1.5 pounds fudge

(b) Leo’s production opportunities:

All time making fudge: 4 pounds; or toffee: 5 pounds.

Opportunity cost of 1 pound fudge = 5/4 = 1.25 pounds toffee

Opportunity cost of 1 pound toffee = 4/5 = 0.8 pounds fudge

Since Ana’s opportunity cost for fudge is lower than Leo’s, she should specialize in fudge production.

Conversely, Leo has a lower opportunity cost for toffee, so he should focus on producing toffee.

7 0
14 days ago
What is the fewest number of points you must plot in order to have examples of all four sets of​ numbers, including at least one
lawyer [4008]

Answer:

Let’s outline the sets:

Integers: The collection of all whole numbers.

Rational: Numbers that can be expressed as the ratio of two integers.

Natural: The collection of positive integers.

Whole numbers: All the values obtainable by repeatedly adding (or subtracting) 1 to a number.

Thus:

2 qualifies as a:

Whole number, because 1 + 1 = 2 (therefore, it’s also an integer).

We can express 2 as 4/2.

As 2 is the fraction of two integers, it is also considered rational.

Since 2 is positive and an integer, it falls into the category of natural numbers.

Therefore, the number 2 exemplifies all four sets.

If we include a negative example, we can consider -3.

-3 is classified as an integer, is not a whole number, and with 9/-3 = -3, thus it is also a rational number.

Now, addressing the questions:

a) Although a single example can be used across all four sets, I have chosen 2 in this instance.

b) Similarly, I demonstrated that 2, as a positive integer, belongs to all four sets, thus:

Positive integers belong to:

The set of integers.

The set of natural numbers.

The set of rational numbers.

The set of whole numbers.

8 0
4 days ago
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