Un sfert din numărul porcinelor din fermă, fiind de rasă superioară, aduc la sfârşitul anului, in medie, o productie de 9.6 kg de carne pe cap de animal. Productia astfel obținută reprezintă jumătate din
totalul productiei de carne,
5 Determinati productia de carne pe cap de animal obținută de la un porc ce nu este de rasă supe-
noara

The meat production per animal obtained from a non-superior breed pig = 4.8 kg of meat per non superior breed of pig.

Producția de carne pe animal obținută de la un porc de rasă non-superioară = 4,8 kg de carne pe o rasă de porc non-superioară.

Step-by-step explanation:

English Translation

A quarter of the number of pigs on the farm, being of superior breed, bring at the end of the year, on average, a production of 9.6 kg of meat per animal. The production thus obtained represents half of total meat production, Determine the meat production per animal obtained from a non-superior breed pig.

Solution

Let the meat production per animal for the non superior breed of pig be y

Let the total number of pigs be x

A quarter of the pigs are superior breed, 0.25x, bring in 9.6 kg of meat per superior breed of pigs.

Total amount of meat produced by the superior breed of pigs = 0.25x × 9.6 = (3.6x) kg

This means that there are three quarter pigs from the less superior breed, 0.75x, making y kg of meat per animal.

Total amount of meat produced by the non superior breed of pigs = 0.75x × y = 0.75xy kg

Total meat production = (3.6x + 0.75xy) kg

It is then given that the total amount of meat from the superior breed is half of the total meat production.

That is,

3.6x = (3.6x + 0.75xy)/2

7.2x = 3.6x + 0.75xy

7.2x - 3.6x = 0.75xy

3.6x = 0.75xy

y = (3.6/0.75) = 4.8 kg

Hence, the meat production per animal obtained from a non-superior breed pig = 4.8 kg of meat per non superior breed of pig.

In Romanian/In limba romana

Permiteți producția de carne pe animal pentru rasa non superioară de porc

Fie numărul total de porci x

Un sfert din rasa superioară, 0,25x, aduce 9,6 kg de carne pentru fiecare rasă superioară de porci.

Un sfert din porci sunt de rasa superioară, 0,25x, aduc 9,6 kg de carne pe rasă superioară de porci.

Cantitatea totală de carne produsă de rasa superioară de porci = 0,25x × 9,6 = (3,6x) kg

Aceasta înseamnă că există trei sferturi de porci de la rasa mai puțin superioară, 0,75x, ceea ce face y kg de carne pe animal.

Cantitatea totală de carne produsă de rasa non superioară de porci = 0,75x × y = (0,75xy) kg

Producția totală de carne = (3,6x + 0,75xy) kg

Se consideră că cantitatea totală de carne de la rasa superioară este jumătate din producția totală de carne.

Acesta este,

3,6x = (3,6x + 0,75xy) / 2

7,2x = 3,6x + 0,75xy

7,2x - 3,6x = 0,75xy

3,6x = 0,75xy

y = (3,6 / 0,75) = 4,8 kg

Prin urmare, producția de carne pe animal obținută de la un porc de rasă non-superioară = 4,8 kg de carne la o rasă de porc ne superioară.

Hope this Helps!!!

Sper că acest lucru vă ajută!!!

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

=====================================

part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

=====================================

part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).