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3 years ago

5 thousand 2 tens 4 ones +4 thousand 3 hundreds 2 ones

Mathematics

2 answers:

adell [148]3 years ago

7 0

5 thousand, 2 tens, 4 ones+ 4 thousand, 3 hundreds, 2 ones= 8 thousand, 3 hundreds, 2 tens, and 6 ones

Digiron [165]3 years ago

3 0

9 thousands 3 hundreds 2 tens 6 ones
or
9326

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How do you do simple algebra

n200080 [17]

Know the order of operations. However, one of the trickiest things about solving an algebra equation as a beginner is knowing where to start. Luckily, there's a specific order for solving these problems: first do any math operations in parentheses, then do exponents, then multiply, then divide, then add, and finally subtract

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3 years ago

For problems 29 - 31 the graph of a quadratic function y=ax^2 + bx + c is shown. Tell whether the discriminant of ax^2 + bx + c

Elan Coil [88]

Answer:

29) discriminant is positive

30) discriminant is 0

31) discriminant is negative

Step-by-step explanation:

the graph of a quadratic function y=ax^2 + bx + c is shown. Tell whether the discriminant of ax^2 + bx + c = 0 is positive, negative, or zero.

In the graph of question number 29 we can see that the graph intersects the x axis at two points

so the equation has 2 solutions.

When the equation has two solution then the discriminant is positive

In the graph of question number 30 we can see that the graph intersects the x axis at only one point

so the equation has only 1 solution.

When the equation has only one solution then the discriminant is equal to 0

In the graph of question number 30 we can see that the graph does not intersects the x axis

so the equation has 2 imaginary solutions.

When the equation has two imaginary solutions then the discriminant is negative

3 0

3 years ago

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:

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

6 0

3 years ago

Can someone please help me with this too

Svetach [21]

Answer:

I have no idea but I suggest making the rule to round the output to the nearest tenth because all the numbers on there have a 9 at the end.

Hope I helped, if I'm right please give brainliest because I cant level up until I get 2 more ;(

<3 Enjoy,

Dea

3 0

3 years ago

Help me with my math please

sergij07 [2.7K]

Step-by-step explanation:

1 _21:00_ , _22:10_, 1:10

2. 1:20

3. 18:00

4. 21:35

5 0

2 years ago

Read 2 more answers