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IRISSAK [1]
3 years ago
6

Find the value of x if y=50. y=5x - 15​

Mathematics
1 answer:
ExtremeBDS [4]3 years ago
3 0

Answer:

x = 13

Step-by-step explanation:

y=50

50 = 5x - 15

50 + 15 = 5x

65 = 5x

65/5 = 5x/5

13 = x

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Write an equation in which the quadratic expression 3x^2+6x-24 equals 0. Show the expression in factored form and explain what y
mr_godi [17]

Answer:

3x² + 6x - 24 = 0

(3x - 6)(x + 4) = 0

x = -4 , x = 2 when y = 0

Step-by-step explanation:

3x² + 6x - 24 = 0

(3x - 6)(x + 4) = 0

Means

3x - 6 = 0 ⇒ 3x = 6 ⇒ x = 6 ÷ 3 = 2

x + 4 = 0 ⇒ x = -4

The Parabola which represents the quadratic equation intersects x-axis at

points (2 , 0) and (-4 , 0)

5 0
3 years ago
Help guys<br> I realy need help <br> I will report you if you give me the wrong answer
telo118 [61]

Answer:

the Answer is 150x + 12100 ≤ 26500

Step-by-step explanation:

The max weight is 26500 so the weight would have to be the same or less than 26500 (≤)

There is already 12100 killograms on the sipping containter so that is why you have + 12100

and you have 150x because x is the number of 150-kilogram crates

and for as many crates you have to multiply by the weight.

7 0
2 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
6 more than 3 times what number squares is 81
Lisa [10]
When you write that equation, you get 3x^2+6=81.

Solving for x:

Subtract by 6 on both sides of the equation to get 3x^2=75.

Then, divide by 3 on both sides of the equation to get x^2=25.

Finally, do the square root on both sides of the equation to get x=5.

The number is 5.
6 0
3 years ago
0.000000000093 in scientific notation
Inga [223]

no because like he said you have to move the decimal places



6 0
3 years ago
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