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Sedbober [7]
3 years ago
5

The equation below describes a parabola. If a is negative, which way does the parabola open? x = ay2

Mathematics
2 answers:
Anastasy [175]3 years ago
8 0
This type of parabola opens either to the left or to the right. The negative makes it open to the left.
Vanyuwa [196]3 years ago
3 0

Answer:

The given parabola opens to the left.

Step-by-step explanation:

We are given,

The equation of the parabola is x=ay^2

If in the parabola,

<em>1. If 'a' is positive, then the parabola opens upwards or right.</em>

<em>2.</em><em> If 'a' is negative, then the parabola opens downwards or left.</em>

Moreover,

<em>3. If x is squared, then the parabola opens upwards or downwards.</em>

<em>4. </em><em>If y is squared, then the parabola opens left or right.</em>

Now, as we are given the parabola having y^2 and 'a' is negative.

Thus, from the points 4 and 2, we get,

The given parabola opens to the left.

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A manufacturing company produces 3 different products A, B, and C. Three types of components, i.e., X, Y, and Z, are used in the
Murljashka [212]

Answer:

Step-by-step explanation:

Using the Excel Formula:

Decision    Variable        Constraint              Constraint

A                     65                          65                         100

B                     80                          80                         80

C                     90                         90                          90

                      14100                    300                        300

= (150 *B3)+(80*B4) +(65*B5)-(100-B3+80-B4+90-B5)*90

Now, we have:

Suppose A, B, C represent the number of units for production A, B, C which is being manufactured

                             A              B                  C                Unit price

Need of X          2                 1                   1                     $20

Need of Y           2                3                  2                    $30

Need of Z           2                2                  3                    $25

Price of  

manufac -      $200          $240            $220      

turing

Now,  for manufacturing one unit of A, we require 2 units of X, 2 units of Y, 2 units of Z are required.  

Thus, the cost or unit of manufacturing of A is:

$20 (2) + $30(2) + $25(2)

$(40 + 60 + 50)

= $150

Also, the market price of A = $200

So, profit = $200 - $150 = $50/ unit of A

Again;

For manufacturing one unit of B, we require 1 unit of X, 3 units of Y, and 2 units of Z are needed and they are purchased at $20, $30, and 425 each.

So, total cost of manufacturing a unit of B is:

= $20(1) + $30(3) + $25(2)

= $(20 + 90+50)

= $160

And the market price of B = $240

Thus, profit = $240- $160  

profit = $80

For manufacturing one unit of C, we have to use 1 unit of X, 2 unit of Y, 3 units of Z are required:

SO, the total cost of manufacturing a unit of C is:

= $20 (1) + $30(2) + $25(3)

= $20 + $60 + $25

= $155

This, the profit = $220 - $155 = $65

However; In manufacturing A units of product A, B unit of product B & C units of product C.

Profit  --> 50A + 80B + 65C

This should be provided there is no penalty for under supply of there is under supply penalty for A, B, C is $40

The current demand is:

100 - A

80 - B

90 - C respectively

So, the total penalty

{(100 - A) + (80 - B) +(90 - C) } + \$40

This should be subtracted from profit.

So, we have to maximize the profit  

Z = 50A + 80B + 65C = {(100 -A) + (80 - B) + (90 - C)};

Subject to constraints;

we have the total units of X purchased can only be less than or equal to 300 due to supplies capacity

Then;

2A + B +C \le 300 due to 2A, B, C units of X are used in manufacturing A, B, C units of products A, B, C respectively.

Next; demand for A, B, C will not exceed 100, 80, 90 units.

Hence;

A \le 100

B \le 80

C \le 90

 

and A, B, C \ge 0 because they are positive quantities

The objective is:

\mathbf{Z = 50A + 80B + 65 C - (100 - A + 80 - B + 90 - C) * 40}

A, B, C \to Decision Varaibles;

Constraint are:

A \le 100 \\ \\  B \le 100 \\ \\ C \le 90 \\ \\2A + B + C \le 300 \\ \\ A,B,C \ge 0

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The pair of equations that has (0, 8) as its solution is: A. equation A and equation C.

<h3>What is the Solution to a System of equations?</h3>

The solution to a system of linear equations that are graphed is the pair of x and y coordinates of the points where two lines intersect.

In the graph given, the lines representing equations A and C intersect at point (0, 8). Therefore, the pair of equation that has a solution of (0, 8) is: A. equation A and equation C.

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Answer:

Step-by-step explanation:

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