Convert 52.68 feet per hour to inches per minute using dimensional analysis

What two numbers can be multiplied to be -150 and added to be 5

pav-90 [236]

Xy = -150
x + y = 5

x + y = 5
x - x + y = -x + 5
y = -x + 5

xy = -150
x(-x + 5) = -150
x(-x) + x(5) = -150
-x² + 5x = -150
-x² + 5x + 150 = 0
-1(x²) - 1(-5x) - 1(-150) = 0
  -1(x² - 5x - 150) = 0
-1 -1
x² - 5x - 150 = 0
x = -(-5) ± √((-5)² - 4(1)(-150))
  2(1)
x = 5 ± √(25 + 600)
2
x = 5 ± √(625)
2
x = 5 ± 25
2
x = 2.5 ± 12.5
x = 2.5 + 12.5 or x = 2.5 - 12.5
x = 15      or        x = -10

x + y = 5
15 + y = 5
- 15 - 15
y = -10
  (x, y) = (15, -10)

  or

x + y = 5
-10 + y = 5
+ 10 + 10
y = 15
  (x, y) = (-10, 15)

The two numbers that add up to 5 and multiply to -150 are 15 and -10.

8 0

3 years ago

A candy company has 150 kg of chocolate-covered nuts and 90 kg of chocolate-covered raisins to be sold as two different mixes. O

Westkost [7]

Answer:

1. 7x + 9.5y

2. 1980

Step-by-step explanation:

In Mix A, there is an equal amount of nuts and raisins.

Let x = kg of nuts = kg of raisins.

In Mix B, there is 3 times as much nuts as raisins.

Let y = kg of raisins, then 3y = kg of nuts.

We have this information:

Nuts Raisins Price

mix A x + x = $ 7.00

mix B 3y + y = $ 9.50

Total: 150 90

Hence, the number of kilograms of each mix should company prepare for the maximum revenue is = 7x + 9.5y

We have these inequalties:

{x≥0y≥0x+3y≤150x+y≤90} (equation 1)

There will be

x+x=2x kg of Mix A at $7.00 per kg.

The revenue from Mix A is:

7(2x)=14x dollars.

There will be

3y+y=4y kg of Mix B at $9.50 per kg.

The revnue from Mix B is:

9.5(4y)=38y dollars.

Hence, the total revenue is:

R=14x+38y (equation 2).

Graph the region determined by [1] and locate its vertices.

You should get:

(0,0),(90,0),(0,50),(60,30)

Test the vertices into [2] and see which vertex produces maximum revenue

(0,0): R=14(0)+38(0)=0 Minimum revenue

(90,0):R=14(90)+38(0)=1260

(0,50): R=14(0)+38(50)=1900

(60,30):R=14(60)+38(30)=1980 Maximum revenue.

3 0

3 years ago

A force of 3 pounds is required to hold a spring stretched 0.6 feet beyond its natural length. how much work (in foot-pounds) is

ioda

The work done (in foot-pounds) in stretching the spring from its natural length to 0.7 feet beyond its natural length is 1.23 foot-pound

<h3>Data obtained from the question</h3>

From the question given above, the following data were obtained:

Force (F) = 3 pounds
Extension (e) = 0.6 feet
Work done (Wd) =?

<h3>How to determine the spring constant</h3>

Force (F) = 3 pounds
Extension (e) = 0.6 feet
Spring constant (K) =?

F = Ke

Divide both sides by e

K = F/ e

K = 3 / 0.6

K = 5 pound/foot

Thus, the spring constant of the spring is 5 pound/foot

<h3>How to determine the work done</h3>

Spring constant (K) = 5 pound/foot
Extention (e) = 0.7 feet
Work done (Wd) =?

Wd = ½Ke²

Wd = ½ × 5 × 0.7²

Wd = 2.5 × 0.49

Wd = 1.23 foot-pound

Therefore, the work done in stretching the spring 0.7 feet is 1.23 foot-pound

Learn more about spring constant:

brainly.com/question/9199238

#SPJ1

5 0

1 year ago

the school store has1200 pendls in stock and sells and sells an average of 25 pencils per day the manager reorders when the numb

forsale [732]

Answer:

Number of pencils in stock = 1200

Average number of pencils sold by the manager per day = 24

Number of pencils that would be sold before reordering = 1200 - 500

= 700

Then

The number of days after which the manager will reorder = 700/24

= 29.16

So the manager has to reorder after about 30 days. Since the answer comes in fraction of more than 29 days, so it has to be 30 days. I hope the procedure is clear for your understanding.

Just to be clear, this answer is not mine, but I remember seeing this question so I just copied someones' answer. Still, you might find this helpful.

Here is the original brainly.com/question/955219

4 0

2 years ago

Which is true if equilibrium is present in a market? Question 5 options: The price of the product will tend to rise. Quantity de

astraxan [27]

On a supply and demand graph you would see that the supply would meet the demand at the central point creating a secure system. However, this could change for the better or worse, compared to how "popular" the item is in the market.

8 0

2 years ago

Read 2 more answers