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LenKa [72]
2 years ago
15

What is the answer?

Mathematics
1 answer:
sertanlavr [38]2 years ago
7 0

Answer:

A. P(3/4)

Step-by-step explanation:

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Name the relationship
Ivanshal [37]

Answer:

A vertical angle!

Step-by-step explanation:

Vertical angles sit across from each other.

4 0
3 years ago
Read 2 more answers
Knife is 3 times cost of spoon<br> 9 spoons and 12 knifes is £82.80<br> what is 1 knife
xenn [34]

Answer:

The cost of spoon is £1.84 and therefore the cost of one knife is £5.52.

Given :

  • A knife is 3 times the cost of the spoon.
  • 9 spoons and 12 knives costs £82.80.

Solution :

  • This question is solve by creating a linear equation and linear equations are nothing but yet another subset of "equations".
  • Linear calculations that requires more than one variable can be done with the help of linear equations.
  • The standard form of a linear equation in one variable is ax + b = 0.

Now let x be the cost of spoon. Than the cost knife is 3x. It is given that 9 spoons and 12 knives costs £82.80.

<h3>9x + 12 x (3x) = 82.809x + 36x = 82.8045x = 82.80x = 1.84</h3>

Therefore the cost of 1 knife = 5.52

Step-by-step explanation:  1.84 x 3 = 5.52

3 0
2 years ago
HELP ⚠️⚠️⚠️⚠️⚠️⚠️⚠️ HELP<br> Please and thank you!
Reika [66]

Answer:

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Step-by-step explanation:

8 0
2 years ago
Complete the following statement: f(3)=_____
victus00 [196]
Using this equation, f(3) = 23.

In order to find the value of f(3), we need to take the f(x) equation and put 3 everywhere we see x. Then we follow the order of operations to solve. So, let's start with the original. 

f(x) = 2x^2 + 5sqrt(x - 2) 

Now place 3 in for each x. 

f(3) = 2(3)^2 + 5sqrt(3 - 2)

Now square the 3.

f(3) = 2(9) + 5 sqrt(3 - 2)

Do the subtraction inside of the parenthesis. 

f(3) = 2(9) + 5sqrt(1)

Take the square root

f(3) = 2(9) + 5(1)

Multiply. 

f(3) = 18 + 5

And add. 

f(3) = 23

4 0
3 years ago
Suppose that a company needs 1, 200,000 items during a year and that preparation for each production run costs $500. Suppose als
MAVERICK [17]

Answer:

The number of items in each production run so that the total costs of production and storage are minimized is 8165 items/run

Step-by-step explanation:

We will use the following variables:

Q = Quantity being ordered

Q* = the optimal order Quantity: the result being sought

D = annual Demand for the item, over the year

P = unit Production cost

S = cost of setting up a production run, regardless of the number of units in the production run (fixed cost per production run)

H = annual cost to Hold one unit

It is important to note which variables are annualized, which are per-order and which are per-unit.

Using the variables, here are the components of the first equation

Total Cost, TC = PC + SC + HC

PC = P x D :  Production Cost = unit Production cost times the annual Demand

SC = (D x S)/Q : Setting up Cost = annual Demand times cost per production setup, divided by the order Quantity (number of units)

HC = (H x Q)/2: Holding Cost = annual unit Holding cost times order Quantity (number of units), divided by 2 (because throughout the year, on average the warehouse is half full).

So TC = PC + SC + HC =  (P x D) + ((D x S)/Q) + ((H x Q)/2) = PD + (DS/Q) + HQ/2

To obtain the optimal order quantity, Q* that minimizes TC, at the minimum TC, dTC/dQ = 0

dTC/dQ = (H/2) – (D x S)/(Q²) = 0

(H/2) – (D x S)/(Q²) = 0

Solving for Q, which is Q* at this point.

(Q*)² = 2DS/H

Q* = √(2DS/H)

D = annual demand for the item = 200000

S = cost of setting up a production run, regardless of the number of units in the production run (fixed cost per production run) = $500

H = annual cost to Hold one unit = $3

Q* = √(2×200000×500/3) = 8164.97 = 8165 items.

3 0
3 years ago
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